{"id":9999,"date":"2019-12-05T12:23:57","date_gmt":"2019-12-05T06:53:57","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9999"},"modified":"2021-07-02T15:43:56","modified_gmt":"2021-07-02T10:13:56","slug":"1st-puc-physics-question-bank-chapter-11","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-question-bank-chapter-11\/","title":{"rendered":"1st PUC Physics Question Bank Chapter 11 Thermal Properties of Matter"},"content":{"rendered":"

You can Download Chapter 11 Thermal Properties of Matter Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Physics Question Bank Chapter 11 Thermal Properties of Matter<\/h2>\n

1st PUC Physics Thermal Properties of Matter Textbook Questions and Answers<\/h3>\n

Question 1.
\nThe triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celcius and Fahrenheit scales.
\nAnswer:
\nTc<\/sub> = T – 273.15
\n\u2234 Tc<\/sub> for neon = 24.57 – 273.15
\n= – 248.58\u00b0C
\nand Tc<\/sub> for carbon dioxide
\n= 216.55 – 273.15
\n= – 56.60\u00b0C
\ntF<\/sub> = \\(\\frac{9}{5}\\) tc<\/sub> + 32
\n\u2234 tF<\/sub> for neon = \\(\\frac{9}{5}\\) \u00d7 (-248.58) + 32
\n= – 415.44\u00b0C
\ntF<\/sub> for carbon dioxide
\n= \\(\\frac{9}{5}\\) \u00d7 (-56.60) + 32
\n= – 69.88\u00b0F<\/p>\n

Question 2.
\nTwo absolute scales A and B have triple points of water defined to be 200A and 350B. What is the relation between TA<\/sub> and TB<\/sub>?
\nAnswer:
\nTriple point of water
\n= 273.16 K = 200A
\n= 350B
\n\u2234 Size of 1K on scale A = \\(\\frac{273.16}{200}\\)
\nand size of 1K on scale B = \\(\\frac{273.16}{350}\\)
\nValue of temperature TA<\/sub> on scale A = \\(\\frac{273.16}{200}\\) TA<\/sub>
\nValue of temperature TB<\/sub> on scale B = \\(\\frac{273.16}{350}\\) TB<\/sub>
\nSince the two temperatures are equal,
\n\\(\\left(\\frac{273.16}{200}\\right) \\mathrm{T}_{\\mathrm{A}}=\\left(\\frac{273.16}{350}\\right) \\mathrm{T}_{\\mathrm{B}}\\)
\n\u21d2 TA<\/sub> = \\(\\frac{4}{7}\\) TB<\/sub><\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nThe electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R0<\/sub> [1 + \u03b1 (T – T0<\/sub>)] The resistance is 101.6\u2126 at the triple point of water 273.16K, and 165.5 \u2126 at the normal melting point of lead (600.5K). What Is the temperature when the resistance is 123.4 \u2126?
\nAnswer:
\nAt T0<\/sub> = 273.16K, R0<\/sub>= 101.6 \u2126
\nAt T1<\/sub> = 600.5K, R1<\/sub> = 165.5 \u2126
\nLet T2<\/sub> be the temperature at which the resistance is R2<\/sub> = 123.4 \u2126
\nR = R0<\/sub> [1 + \u03b1 (T – T0<\/sub>)]
\n\u21d2 T – T0<\/sub> = \\(\\frac{\\mathrm{R}-\\mathrm{R}_{0}}{\\alpha \\mathrm{R}_{0}}\\)
\n\u2234 T1<\/sub> – T0<\/sub> = \\(\\frac{\\mathrm{R}_{1}-\\mathrm{R}_{0}}{\\alpha \\mathrm{R}_{0}}\\)\u2192 (1)
\nT2<\/sub> – T0<\/sub> = \\(\\frac{\\mathrm{R}_{2}-\\mathrm{R}_{0}}{\\alpha \\mathrm{R}_{0}}\\)\u2192 (2)
\nDividing (2) by (1) gives,
\n\\(\\frac{\\mathrm{T}_{2}-\\mathrm{T}_{0}}{\\mathrm{T}_{1}-\\mathrm{T}_{0}}=\\frac{\\mathrm{R}_{2}-\\mathrm{R}_{0}}{\\mathrm{R}_{1}-\\mathrm{R}_{0}}\\)
\n\\(\\mathrm{T}_{2}=\\mathrm{T}_{0}+\\left(\\mathrm{T}_{1}-\\mathrm{T}_{0}\\right)\\left(\\frac{\\mathrm{R}_{2}-\\mathrm{R}_{0}}{\\mathrm{R}_{1}-\\mathrm{R}_{0}}\\right)\\)
\n= 293.16 + (600.5-273.16) \\(\\left(\\frac{123.4-101.6}{165.5-101.6}\\right)\\)
\n\u21d2 T2<\/sub> = 384.83 K<\/p>\n

Question 4.
\nAnswer the following:
\n1. The triple point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the
\nboiling point of water as standard fixed points (as was originally done in the Celcius scale)?<\/p>\n

2. There were two fixed points in the original Celcius scale as mentioned above which were assigned the number 0\u00b0C and 100\u00b0C respectively. On the absolute scale, one of the fixed points is the triple point of water, which on the Kelvin absolute scale is assigned the number 273.16K. What is the other fixed point on this (Kelvin) scale?<\/p>\n

3. The absolute temperature (Kelvin scale) T is related to the temperature te on the Celcius scale by tc<\/sub> = T – 273.15 Why do we have 273.15 in this relation and not 273.16?<\/p>\n

4. What is the temperature of the triple point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
\nAnswer:
\n1. The triple point is used as a standard because of the unique conditions of temperature and pressure at this point. The melting point of ice and boiling point of water depend on pressure. So they cannot be used as reference points.<\/p>\n

2. The other fixed point is absolute zero, the theoretically lowest possible temperature.<\/p>\n

3. We have 273.15 in the relation and not 273.16 because the triple point of water is 0.01 \u00b0C and not 0\u00b0C.
\nIn other words, 0.01 \u00b0C \u2192 273.16 K
\n\u21d2 0\u00b0C \u2192 273.15 K
\n\u2234 tc<\/sub> = T – 273.15<\/p>\n

4. The unit interval size in the Fahrenheit scale = \\(\\frac{212-32}{100}\\) =1.8
\n\u2234 The triple point of water = 273.16 \u00d7 1.8
\n= 491.69<\/p>\n

Question 5.
\nTwo ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:
\n\"1st
\n1. What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
\n2. What do you think Is the reason behind the slight difference in the answers of thermometers A and B? (The thermometers are not faulty) What further procedure is needed In the experiment to reduce the discrepancy between the two readings?
\nAnswer:
\n1. Let P be the pressure in the pressure thermometer and T be the corresponding absolute temperature.
\nLet P0<\/sub> and T0<\/sub> be the values for the triple point of water.
\nThen \\(\\frac{T}{T_{0}}=\\frac{P}{P_{0}}\\)
\n\\(\\therefore \\frac{T_{A}}{273.16}=\\frac{1.797 \\times 10^{5}}{1.250 \\times 10^{5}}\\)
\n\u21d2 TA<\/sub> = 392.69 K
\n\\(\\frac{T_{B}}{273.16}=\\frac{0.287 \\times 10^{5}}{0.200 \\times 10^{5}}\\)
\n\u21d2 TB<\/sub>= 391.98 K<\/p>\n

2. The discrepancy is due to the non\u00acideal behaviour of the gases. To reduce this discrepancy, measurements at lower pressures should be taken.<\/p>\n

Question 6.
\nA steel tape 1m long Is correctly calibrated for a temperature of 27.0\u00b0C. The length of a steel rod measured by this tape Is found to be 63.0 cm on a hot day when the temperature is 45.0\u00b0C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0\u00b0 C? Coefficient of linear expansion of steel = 1.20 \u00d7 10-5<\/sup>K-1<\/sup>.
\nAnswer:
\nl0<\/sub> = 63 cm
\n\u2206T = 45\u00b0C – 27\u00b0C .
\n= 18\u00b0C = 18 K
\n\u03b1 = 1.2 \u00d7 10-5<\/sup>K-1<\/sup>
\nl = l0<\/sub> (1 + \u03b1 \u2206T)
\n= 63(1 + 1.2 \u00d7 10-5<\/sup> \u00d7 18)
\nl = 63.0136 cm.
\n\u2234 Actual length of the steel rod at 45\u00b0C =63.0136 cm
\nAt 27\u00b0C, the steel tape is correctly calibrated.
\n\u2234 the length of the steel rod = 63.0 cm<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nA large steel wheel Is to be fitted on to a shaft of the same material. At 27\u00b0C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole In the wheel is 8.69 cm. The shaft Is cooled using \u2018dry ice\u2019. At what temperature of the shaft does the wheel slip on the shaft? Assume the coefficient of linear expansion of the steel to be constant over the required temperature range: \u03b1steel<\/sub> = 1.2 \u00d7 10-5<\/sup>K-1<\/sup>.
\nAnswer:
\nGiven \u03b1steel<\/sub> = 1.2 \u00d7 10-5<\/sup>K-1<\/sup>
\nl1<\/sub> = 8.70 cm
\nl2<\/sub> = 8.69 cm (because the shaft is cooled to help it slip on the wheel)
\nT1<\/sub> = 27\u00b0C = 300 K
\nT2<\/sub> =?
\nl2<\/sub> = l1<\/sub> [1 + \u03b1steel<\/sub>(T2<\/sub> – T1<\/sub>)]
\n\u21d2 8.69 = 8.70 [1 + 1.2 \u00d7 10-5<\/sup>(T2<\/sub> – 300)]
\n\u21d2 1 + 1.2 \u00d7 10-5<\/sup>(T2<\/sub> – 300) = \\(\\frac{8.69}{8.70}-1\\)
\n\u21d2 T2<\/sub> = 300 + \\(\\frac{-0.01}{8.70} \\times \\frac{1}{1.2 \\times 10^{-5}}\\)
\n= 204.21 K
\n= – 68.94\u00b0C
\n= – 69\u00b0C
\n\u2234 The required temperature = – 69\u00b0C<\/p>\n

Question 8.
\nA hole drilled in a copper sheet. The diameter of the hole is 4.24cm at 27.0\u00b0C. What is the change in the diameter of the hole when the sheet is heated to 227\u00b0 C? Coefficient of linear expansion of copper = 1.70 \u00d7 10-5<\/sup>K-1<\/sup>.
\nAnswer:
\nGiven \u03b1 = 1.70 \u00d7 10-5<\/sup>K-1<\/sup>
\n\u03b2 = 2\u03b1 = 3.40 \u00d7 10-5<\/sup>K-1<\/sup>
\nd1<\/sub> = 4.24 cm
\nInitial area A1<\/sub> at 27\u00b0C = 300K = T1<\/sub>
\nis \\(\\pi\\left(\\frac{\\mathrm{d}_{1}}{2}\\right)^{2}=\\pi\\left(\\frac{4.24}{2}\\right)^{2}\\)
\n\u21d2 A1<\/sub> = 4.4944 \u03c0 cm2<\/sup>
\nArea A2<\/sub> at 227\u00b0C = 500K = T2<\/sub> is
\n\u22061<\/sub> [1 + \u03b2(T2<\/sub> – T1<\/sub>)]
\ni.e. A2<\/sub> = 4.4944 \u03c0 [1 + 3.40 \u00d7 10-5<\/sup> (500 – 300)]
\nA2<\/sub> = 4.5250 \u03c0 cm2<\/sup>
\n\u21d2 \\(\\pi\\left(\\frac{\\mathrm{d}_{2}}{2}\\right)^{2}\\) = 4.5250 \u03c0 cm2<\/sup>
\n\u21d2 d2<\/sub> = \\(\\sqrt{18.100}\\) =4.2544 cm
\n\u2234 Change in diameter = d2<\/sub> – d1<\/sub>
\n= 4.2544 – 4.25 = 0.0144 cm<\/p>\n

Question 9.
\nA brass wire 1.8 m long at 27\u00b0C Is held taut with little tension between two rigid supports. If the wire Is cooled to a temperature of -39\u00b0C, what Is the tension developed In the wire, if its diameter is 2.0 mm?
\nCoefficient of linear expansion of brass = 2.0 \u00d7 10-5<\/sup> K-1<\/sup>, Young\u2019s modulus Of brass = 0.91 \u00d7 1011<\/sup> Pa.
\nAnswer:
\nl = 1.8 m,
\nT1<\/sub> =27\u00b0C = 300 K,
\nT2<\/sub> = – 39\u00b0C =234 K,
\nd = 2mm = 2 \u00d7 10-3<\/sup>
\n\u03b1brass<\/sub> = 2.0 \u00d7 10-5<\/sup> K-1<\/sup>,
\nYbrass<\/sub> = 0.91 \u00d7 1011<\/sup> Pa
\n\\(\\frac{\\Delta \\ell}{\\ell}=\\alpha \\Delta \\mathrm{t}\\)
\n\\(Y=\\frac{\\Delta F \/ A}{\\Delta \\ell \/ \\ell}=Y\\left(\\frac{\\Delta \\ell}{\\ell}\\right) A\\)
\n= Y(\u03b1 \u2206T) \\(\\pi\\left(\\frac{\\mathrm{d}}{2}\\right)^{2}\\)
\n= 0.91 \u00d7 1011<\/sup> \u00d7 2.0 \u00d7 10-5<\/sup> \u03b1 (234 – 300) \u00d7 \u03c0 \u00d7 \\(\\left(\\frac{2 \\times 10^{-3}}{2}\\right)^{2}\\)
\n= – 377.37 N
\nTension developed \u2206F = 380N (positive value, rounded to 2 significant figures)<\/p>\n

Question 10.
\nA brass rod of length 50cm and diameter 3.0 mm is Joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250\u00b0C, if the original lengths are at 40.0\u00b0C? Is there a \u2018thermal stress\u2019 developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass = 2.0 \u00d7 10-5<\/sup>K-1<\/sup>, steel = 1.2 \u00d7 10-5<\/sup>K-1<\/sup>).
\nAnswer:
\nGiven:\u00a0 lbrass<\/sub> = 50cm, lsteel<\/sub> = 50cm
\nT1<\/sub> = 40\u00b0C = 313K
\nT2<\/sub> = 250\u00b0C = 523 K
\n\u03b1brass<\/sub> = 2.0 \u00d7 10-5<\/sup>K-1<\/sup>
\n\u03b1steel<\/sub> = 1.2 \u00d7 10-5<\/sup>K-1<\/sup>
\n\u2206lbrass<\/sub> = lbrass<\/sub> \u00d7 \u03b1brass<\/sub> \u00d7 \u2206T
\n= 50 \u00d7 2.0 \u00d7 10-5<\/sup> \u00d7 (523 – 313)
\n\u2206lbrass<\/sub> = 0.21 cm
\n\u2206lsteel<\/sub> = lsteel<\/sub> \u00d7 \u03b1steel<\/sub> \u00d7 \u2206T
\n= 50 \u00d7 1.2 \u00d7 10-5<\/sup> \u00d7 (523 – 313)
\n= 0.126 cm
\n\u2206lsteel<\/sub> = 0.13cm
\n(round off to two significant figures) Since the rods are free to expand, no thermal stress is developed at the junction.<\/p>\n

Question 11.
\nThe coefficient of volume expansion of glycerin is 49 \u00d7 10-5<\/sup>K-1<\/sup>. What is the fractional change in its density for a 30\u00b0C rise In temperature?
\nAnswer:
\nGiven, rglycerin<\/sub> = 49 \u00d7 10-5<\/sup>K-1<\/sup>
\n\u2206T = 30\u00b0C = 30 K
\nLet the volume be V1<\/sub> at temperature and V2<\/sub> at temperature T2<\/sub>
\nUsing V2<\/sub> = V1<\/sub>(1 + r \u2206T), we get
\n\\(\\frac{V_{1}}{V_{2}}=\\frac{1}{1+r \\Delta T}\\)
\n\\(=\\frac{1}{1+49 \\times 10^{-5} \\times 30}=0.9855\\)
\nFractional change in density
\n\"1st
\n= 0.9855 – 1
\n= – 0.0145
\n= – 1.5 \u00d7 10-2<\/sup>
\n(The negative sign denotes a decrease in density).<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nA10 kW drilling machine is used to drill a bore In a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the blocks in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.91 J g-1<\/sup> K-1<\/sup>
\nAnswer:
\nGiven,
\nP = 10kW,
\nt = 2.5 minutes = 150s,
\nm = 8 kg,
\nsaluminium<\/sub> = 0.91 Jg-1<\/sup> K-1<\/sup> = 910 Jkg-1<\/sup> K-1<\/sup>
\nEnergy transferred to the aluminium
\n= 50% of p \u00d7 t
\nThis energy heats up the aluminium.
\nSo, \\(\\frac{50}{100}\\) \u00d7 pt = m saluminum<\/sub> \u2206T
\n\u21d2 \\(\\frac{1}{2}\\) \u00d7 10 \u00d7 103<\/sup> \u00d7 150 = 8 \u00d7 910 \u00d7 \u2206T
\n\u21d2 \u2206T = 103 K
\nor \u2206T = 103\u00b0C<\/p>\n

Question 13.
\nA copper block of mass 2.5kg is heated in a furnace to a temperature of 500\u00b0<\/sub>C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 Jg-1<\/sup> K-1<\/sup> heat of fusion of water = 335 Jg-1<\/sup>)
\nAnswer:
\nGiven mcopper<\/sub> = 2.5 kg,
\nT1<\/sub>= 500\u00b0C,
\nT2<\/sub> = 0\u00b0C,
\nScopper<\/sub> = 0-39 Jg-1<\/sup> K-1<\/sup>
\nLice<\/sub> = 335 Jg-1<\/sup>
\nLet the mass of ice that melts be mice<\/sub>
\nHeat dissipated by the copper block = Heat absorbed by the ice.
\n\u21d2 mcopper<\/sub> Scopper<\/sub> \u2206T = mice<\/sub> Lice<\/sub>
\n\u21d2 2500 \u00d7 0.39 \u00d7 (500 – 0) = mice<\/sub> \u00d7 335
\n\u21d2 mice<\/sub> = 1455 g
\n\u21d2 mice<\/sub> = 1.5 kg
\n(rounding off to 2 significant figures)<\/p>\n

Question 14.
\nIn an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150\u00b0C Is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3<\/sup> of water at 27\u00b0C. The final temperature is 40\u00b0C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value of the specific heat of the metal?
\nAnswer:
\nGiven, mass of metal block,
\nm1<\/sub> = 0.20 kg
\nInitial temperature of the metal block,
\nT1<\/sub> = 150\u00b0C
\nInitial temperature of the calorimeter and water,
\nT2<\/sub> = 27\u00b0<\/sub>C
\nFinal temperature of the mixture,
\nT3<\/sub> = 40\u00b0C
\nVolume of water in the calorimeter,
\nV =150 cm3<\/sup>
\n= 150 \u00d7 10-6<\/sup> m3<\/sup>
\nMass of the water = Density \u00d7 Volume
\n= 1000 \u00d7 150 \u00d7 10-6<\/sup>
\n= 150 \u00d7 10-3<\/sup> kg
\nLet s1<\/sub> be the specific heat of the metal.
\ns2<\/sub> is the specific heat of water.
\nWater equivalent of the calorimeter,
\nW = 0.025 kg
\nHeat lost by the metal block = Heat gained by the calorimeter and water.
\n\u21d2 m1<\/sub> s1<\/sub> \u2206T1<\/sub> = (m2<\/sub> + W) s2<\/sub> \u2206T2<\/sub>
\n\u21d2 0.2 \u00d7 s1<\/sub> \u00d7 (150 – 40)
\n= (150 \u00d7 10-3<\/sup> + 0.025) \u00d7 4200 \u00d7 (40 – 27)
\n\u21d2 s1<\/sub> = 434.3 J kg-1<\/sup>K-1<\/sup>
\n= 430 J kg-1<\/sup>K-1<\/sup>
\nor 0.43 Jg-1<\/sup>K-1<\/sup>
\nIf the heat losses to the surroundings are not negligible, the left-hand side of the expression would have had an extra term added to it to account for this loss. Upon solving for s2<\/sub> we would get a smaller value than the actual value.<\/p>\n

Question 15.
\nGiven below are observations on molar specific heats at room temperature of some common gases.
\nGas\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Molar specific heat (cv) (cal mol\u20191 K\u20191)
\nHydrogen\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a04.87
\nNitrogen\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a04.97
\nOxygen\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a05.02
\nNitric oxide\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a04.99
\nCarbon monoxide\u00a0 \u00a05.01
\nChlorine\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 6.17
\nThe measured molar specific heats of these gases are markedly different from those for monoatomic gases. Typically, the molar specific heat of a monoatomic gas is 2.92 cal\/mol K. Explain this difference. What can you Infer from the somewhat larger (than the rest) value for chlorine?
\nAnswer:
\nAll the gases mentioned above are diatomic, and hence have other degrees of freedom in addition to the translational degrees of freedom. To raise the temperature of the gas, heat has to be supplied to increase the average energy of all the modes.<\/p>\n

Consequently, the molar specific heats for diatomic gases is higher than that for monoatomic gases. The higher value of the molar specific heat of chlorine indicates that at room temperature, vibrational modes are also present In addition to the rotational modes of freedom.<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nAnswer the following questions based on the P-T diagram of carbon dioxide:
\n1. At what temperature and pressure can the solid, liquid and vapour phases of Co2<\/sub>, exist in equilibrium?
\n2. what is the effect of a decrease of pressure on the fusion and boiling point of Co2<\/sub>?
\n3. What are the critical temperature and pressure for Co2<\/sub>? What is their significance?
\n4. Is Co2<\/sub> solid, liquid or gas at<\/p>\n

    \n
  • – 70\u00b0C under 1 atm,<\/li>\n
  • – 60\u00b0C under 10 atm,<\/li>\n
  • 15\u00b0C under 56 atm?<\/li>\n<\/ul>\n

    Answer:
    \n\"1st
    \n1. At -56.6\u00b0C and 5.11 atm, which is the triple point of Co2<\/sub>, all 3 .phases can coexist in equilibrium.
    \n2. From the fusion and vaporisation curves, we.see that reducing the pressure lowers the fusion and boiling point.
    \n3. The critical temperature and pressure are 31.1\u00b0C and 73.0 atm respectively. Above this temperature, Co2<\/sub> will not liquefy even if compressed to high pressures.
    \n4.<\/p>\n

      \n
    • Gas<\/li>\n
    • Solid<\/li>\n
    • Liquid<\/li>\n<\/ul>\n

      Question 17.
      \nAnswer the following questions based on the P-T phase diagram of Co2<\/sub>:<\/p>\n

        \n
      1. Co2<\/sub> at 1 atm pressure and temperature – 60\u00b0C is compressed isothermally. Does it go through a liquid phase?<\/li>\n
      2. What happens when Co2<\/sub> at 4 atm pressure is cooled from room temperature at constant pressure?<\/li>\n
      3. Describe qualitatively the changes In a given mass of solid Co2<\/sub> at 10 atm pressure and temperature -65\u00b0C as it is heated up to room temperature at constant pressure.<\/li>\n
      4. Co2<\/sub> is heated to a temperature 700C and compressed isothermally. What changes in its properties do you expect to observe?<\/li>\n<\/ol>\n

        Answer:
        \n\"1st<\/p>\n

          \n
        1. When subjected to isothermal compression, Co2<\/sub> will condense to solid directly without passing through the liquid phase.<\/li>\n
        2. When cooled at constant pressure, Co2<\/sub>\u00a0will condense directly to solid phase without passing through the liquid phase,<\/li>\n
        3. As the solid Co2<\/sub> is heated up, it first turns to liquid phase and then solid phase. The fusion and boiling points can be obtained from finding out the points of intersection between the horizontal 10 atm line on the P-T diagram with the fusion and vaporisation curves.<\/li>\n
        4. Co2<\/sub> will not exhibit any clear transition to the liquid phase but will deviate more and more from ideal behaviour ps the pressure increases.<\/li>\n<\/ol>\n

          Question 18.
          \nA child running a temperature of 101\u00b0F is given an antipyrin (i.e., a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98\u00b0F in 20 min, what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of the human body is approximately the same as that of water, and latent heat of evaporation of water \u2018at that temperature is about 580 cal g-1<\/sup>.
          \nAnswer:
          \nMass of the child, m = 30 kg
          \nChange in temperature \u2206T ,
          \n= 101\u00b0F – 98\u00b0F
          \n= 3\u00b0F
          \n= 3 \u00d7 5\/9 = 5\/3\u00b0c
          \nSpecific heat of water,
          \ns = 4.2 \u00d7 103 J kg-1<\/sup>\u00b0C-1<\/sup>
          \nLatent heat of vaporisation,
          \nL = 580 cal g-1<\/sup>
          \n= 580 \u00d7 4.2 \u00d7 103<\/sup> J kg-1<\/sup>
          \nTime taken for the sweat to evaporate,
          \nt = 20 min.
          \nLet m1<\/sub> be the mass of sweat that evaporates.
          \nNow, heat lost by the child during evaporation
          \n= Heat needed for the sweat to evaporate.
          \n\u21d2 ms \u2206T = m1<\/sub>L
          \n\u21d2 30 \u00d7 4.2 \u00d7 103<\/sup> \u00d7 5\/3
          \n= m1<\/sub> \u00d7 580 \u00d7 4.2 \u00d7 103<\/sup>
          \n\u21d2 m1<\/sub> = 0.0862 kg
          \nm1<\/sub> = 86.2 g
          \nRate of extra evaporation caused by the drug = Rate of evaporation of sweat
          \n\\(=\\frac{m_{1}}{t}=\\frac{86.2 g}{20 m i n}\\)
          \n= 4.3 g min-1<\/sup><\/p>\n

          Question 19.
          \nA ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food In summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg pt Ice is put in the box, estimate the amount of Ice remaining after 6h. The outside temperature is 45\u00b0C, and co-efficient of thermal conductivity of thermacole is 0.01 J s-1<\/sup> m-1<\/sup>K-1<\/sup>. [Heat of fusion of water = 835 \u00d7 103<\/sup> J kg-1<\/sup>]
          \nAnswer:
          \nTotal surface area of the cube,
          \nA = 6 \u00d7 (30 cm)2<\/sup>
          \n= 0.54 m2<\/sup>
          \nMass of ice in the box M = 4 kg
          \nThickness of the box, L = 5 cm = 0.05 m
          \nThermal conductivity of thermacole,
          \nk = 0.01 Js-1<\/sup>m-1<\/sup>K-1<\/sup>.
          \nOutside temperature T1<\/sub> = 45\u00b0C
          \nTemperature of ice T2<\/sub> = 0\u00b0C
          \nLet the mass of ice that melts be m kg.
          \nHeat of fusion of water,
          \nLf ice<\/sub> = 335 \u00d7 103<\/sup> J kg-1<\/sup>
          \nTime t = 6h = 6 \u00d7 3600 s = 21600 s
          \nHeat gained by ice during fusion = Heat conducted through the walls of the box
          \n\u21d2 m Lf, ice<\/sub> = \\(\\frac{\\mathrm{KA}\\left(\\mathrm{T}_{1}-\\mathrm{T}_{2}\\right) \\mathrm{t}}{\\mathrm{L}}\\)
          \n\u21d2 m \u00d7 335 \u00d7 103<\/sup>
          \n\\(=\\frac{0.01 \\times 0.54 \\times(45-0)}{0.05} \\times 21600\\)
          \n\u21d2 m = 0.31 kg
          \n\u2234 Mass of ice left in the box = M – m
          \n= 4 – 0.31
          \n= 3.69
          \n= 3.7 kg.<\/p>\n

          \"KSEEB<\/p>\n

          Question 20.
          \nA brass boiler has a base area of 0.15 m2<\/sup> and thickness 1.0 cm. It bolls water at the rate of 6.0 kg\/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s-1<\/sup>m-1<\/sup>K-1<\/sup>; Heat of vaporisation of water = 2256 \u00d7 103<\/sup> J kg-1<\/sup>.
          \nAnswer:
          \nLet the temperature of the flame be T1<\/sub>
          \nT2<\/sub> = boiling point of water = 100\u00b0C
          \nThermal conductivity of brass,
          \nk = 109 J s-1<\/sup>m-1<\/sup>K-1<\/sup>
          \nBase area of the boiler, A = 0.15 m2<\/sup>
          \nThickness of the boiler, L = 1 cm = 0.01 m
          \n\u21d2 Heat of vaporisation of water
          \nLsteam<\/sub> = 2256 \u00d7 103 J kg-1<\/sup>
          \nRate of boiling water
          \n= 6 kg min-1<\/sup> = \\(\\frac{6}{60}\\) kg s-1<\/sup>
          \n= 0.1 kg s-1<\/sup>
          \n\u2234 Power consumed = 0.1 \u00d7 Lsteam<\/sub>
          \n= 0.1 \u00d7 2256 \u00d7 103<\/sup>W
          \n= 2256 \u00d7 102<\/sup>W
          \n\u2234 Powe transmitted through the boiler
          \n= 2256 \u00d7 102<\/sup> W
          \n\\(\\Rightarrow \\frac{\\mathrm{KA}\\left(\\mathrm{T}_{1}-\\mathrm{T}_{2}\\right)}{\\mathrm{L}}\\) = 2256 \u00d7 102<\/sup>
          \n\u21d2 \\(\\frac{109 \\times 0.15 \\times\\left(T_{1}-100\\right)}{0.01}\\) = 2256 \u00d7 102<\/sup>
          \n\u21d2 T1<\/sub> – 100 = 138
          \n\u21d2 T1<\/sub> = 238\u00b0C
          \n\u2234 Temperature of the flame = 238\u00b0C<\/p>\n

          Question 21.
          \nExplain why:
          \n1. a body with large reflectivity Is a poor emitter.<\/p>\n

          2. a brass tumbler feels much colder than a wooden tray on a chilly day.<\/p>\n

          3. an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open but gives a correct value for the temperature when the same piece is in the furnace.<\/p>\n

          4. The earth without its atmosphere would be inhospitably cold.<\/p>\n

          5. heating systems based on circulation of steam are more efficient in warming a building than those based on the circulation of hot water.
          \nAnswer:
          \n1. We know that a + r + 1 = 1, where \u2018a\u2019 is absorbance, \u2018r\u2019 is reflectance and \u2018t\u2019 is transmittance or emittance. According to Kirchhoff\u2019s law, emittance \u221d absorbance, i.e., good absorbers are also good emitters and so are poor reflectors. Therefore, if reflectivity is large (\u2018r\u2019 is large) then \u2018a\u2019 is small and hence emittance is smaller (the object behaves as a poor emitter).<\/p>\n

          2. The coefficient of thermal conductivity of brass is higher than that of wood. When a brass tumbler is touched, heat quickly flows from human body to the tumbler and so the tumbler appears colder. But in the case of the wooden tray heat does not flow from the human body to the wooden tray, and so it feels relatively hotter.<\/p>\n

          3. Let the temperature of the hot iron in the furnace be\u2018T\u2019 and that of the external environment be \u2018T0<\/sub>\u2019. According to Stefan\u2019s Law, heat energy radiated per second E = \u03c3T4<\/sup> in the open. Clearly E > E’. Hence the low readings of the optical pyrometer when the iron piece is in the open.<\/p>\n

          4. The earth\u2019s atmosphere acts like an insulating blanket around it and does not allow heat to escape out but reflects it back to the earth. If it were absent, the heat would have escaped and the planet would be very cold.<\/p>\n

          5. Steam has a higher heat capacity (due to the latent heat) than boiling water. Therefore steam-based water heating systems are more effective.<\/p>\n

          \"KSEEB<\/p>\n

          Question 22.
          \nA body cools from 80\u00b0C to 50\u00b0C in 5 minutes. Calculate the time it takes to cool from 60\u00b0C to 30\u00b0C. The temperature of the surroundings is 20\u00b0C.
          \nAnswer:
          \nThe average temperature of 80\u00b0 C and 50\u00b0 C is 65\u00b0 C, which is 45\u00b0 C above the room temperature. Under these conditions, the body cools 30\u00b0C in 5 minutes. Using the equation.
          \n\\(\\frac{\\text { Change in temperature }}{\\text { Time }}=\\mathrm{KT}\\)
          \nwe have \\(\\frac{30^{\\circ} \\mathrm{C}}{5 \\mathrm{min}}\\) = K (45\u00b0 C) ……(1)
          \nThe average of 60\u00b0 C and 30\u00b0 C is 45\u00b0 C which is 25\u00b0 C above the room temperature. K is the same for this situation as the original.
          \nHence \\(\\frac{30^{\\circ} \\mathrm{C}}{\\mathrm{t}}\\) = K (25\u00b0 C) ……(2)
          \nDivide (1) by (2), we get,
          \n\"1st
          \n\u2234 Time taken = 9 minutes.<\/p>\n

          1st PUC Physics Thermal Properties of Matter One Mark Questions and Answers<\/h3>\n

          Question 1.
          \nDefine coefficient of thermal conductivity.
          \nAnswer:
          \nCoefficient of thermal conductivity of a material is defined as the quantity of heat conducted per second through a unit area of a slab of unit thickness when the temperature difference between its ends is 1K.<\/p>\n

          Question 2.
          \nWhat is thermal radiation?
          \nAnswer:
          \nHeat radiation from a body due to its temperature is called thermal radiation.<\/p>\n

          Question 3.
          \nWhat is a perfect black body?
          \nAnswer:
          \nA body, which absorbs all the radiation incident on it is called a perfect black body.<\/p>\n

          Question 4.
          \nDefine emissivity of a surface.
          \nAnswer:
          \nEmissivity of a surface is defined as the ratio of the quantity of thermal radiation emitted per sec per unit area of the surface to the quantity of thermal radiation emitted per sec per unit area of a perfect black body under identical conditions.<\/p>\n

          \"KSEEB<\/p>\n

          Question 5.
          \nDefine absorptivity of a surface.
          \nAnswer:
          \nAbsorptivity of a surface is defined as the ratio of the quantity of thermal radiation absorbed per sec to the quantity of radiation incident on it per second.<\/p>\n

          Question 6.
          \nWhat is the unit of coefficient of thermal conductivity?
          \nAnswer:
          \nWm-1<\/sup>K-1<\/sup><\/p>\n

          Question 7.
          \nWhat is the dimensional formula of coefficient of thermal conductivity?
          \nAnswer:
          \nMLT-3<\/sup> \u03b8-1<\/sup><\/p>\n

          Question 8.
          \nWhat happens to the heat supplied to a good conductor when it is under steady-state?
          \nAnswer:
          \nHeat is entirely conducted from the hot end to the cold end.<\/p>\n

          Question 9.
          \nWhy is coal tar heated before use?
          \nAnswer:
          \nCoal tar is heated to reduce its viscosity.<\/p>\n

          \"KSEEB<\/p>\n

          Question 10.
          \nWhat is the value of emissivity of a perfect black body?
          \nAnswer:
          \nEmissivity of a perfect black body is unity.<\/p>\n

          Question 11.
          \nGive dimensions of solar constant and water equivalent.
          \nAnswer:
          \nSolar constant: [MT-3<\/sup>]
          \nWater equivalent: [M]<\/p>\n

          1st PUC Physics Thermal Properties of Matter Two Marks Questions and Answers<\/h3>\n

          Question 1.
          \nExplain three methods of transfer of heat.
          \nAnswer:<\/p>\n

          1. Conduction:
          \nIt is a process of transmission of heat in solids, without the actual movement of the particles.<\/p>\n

          2. Convection:
          \nIt is a process of transmission of heat in fluids (liquids and gases) with the actual movement of particles.<\/p>\n

          3. Radiation:
          \nIt is a process of transmission of heat without any medium.<\/p>\n

          Question 2.
          \nExplain steady-state and temperature gradient.
          \nAnswer:
          \nWhen a solid is heated at one end, heat starts flowing from hot end to cold end. Heat supplied to any element of the solid is partly used in heating it, partly conducted through it and rest is radiated from its surface. After a certain stage, there is no variation of temperature with time and heat supplied is entirely conducted and radiated.<\/p>\n

          Such a state is called steady state. Under steady-state, the temperature at different parts of the solid remains constant and is independent of time. The ratio of the difference in temperature to the distance between two points in a heated body under steady state is called the temperature gradient.<\/p>\n

          \"KSEEB<\/p>\n

          Question 3.
          \nMention the properties of thermal radiation.
          \nAnswer:
          \nProperties of thermal radiation:<\/p>\n

            \n
          1. It travels through a vacuum with the velocity of light.<\/li>\n
          2. It travels in a straight line.<\/li>\n
          3. It undergoes reflection, refraction and total internal reflection.<\/li>\n
          4. It exhibits the phenomenon of interference, diffraction and polarisation.<\/li>\n
          5. Intensity of radiation decreases with increase in distance.<\/li>\n<\/ol>\n

            Question 4.
            \nExplain Ferry\u2019s black body.
            \nAnswer:
            \n\"1st
            \nFerry\u2019s black body consists of a hollow double-walled sphere with a fine hole and a pointed projection in front of the hole. The space between the two walls is evacuated to reduce any loss of radiation due to convection. The inner walls are coated with lamp black. Any radiation falling on the hole is completely absorbed due to multiple reflections, hence the hole acts as a black body.<\/p>\n

            Question 5.
            \nState and explain Kirchhoff\u2019s law.
            \nAnswer:
            \nAt a given temperature, the ratio of the emissive power to the absorptive power for all bodies is a constant and is equal to the emissive power of a perfect black body.
            \nIf e is the emissive power and a is the absorptive power then, \\(\\frac{\\mathrm{e}}{\\mathrm{a}}\\) = constant\u00a0 =1.
            \n\u2234 e = a(\u2235 emissive power of a perfectly black body = 1)
            \nThus, the emissive power of a body is equal to its absorptive power, which means good emitter of thermal radiation is also a good absorber of heat.<\/p>\n

            \"KSEEB<\/p>\n

            Question 6.
            \nState and explain Newton\u2019s law of cooling.
            \nAnswer:
            \nThe rate of cooling of a body is directly proportional to difference in temperature of the body and the surrounding. If the temperature of a body
            \ndecreases from \u03b81<\/sub> to \u03b82<\/sub> in t seconds, then rate of cooling is, \\(\\frac{\\theta_{1}-\\theta_{2}}{t}\\)
            \nAverage temperature of the body \u03b8 = \\(\\frac{\\theta_{1}+\\theta_{2}}{2}\\)
            \nAccording to Newton\u2019s law,
            \n\\(\\frac{\\theta_{1}-\\theta_{2}}{t} \\propto\\left(\\theta-\\theta_{0}\\right)\\)
            \n\u03b80<\/sub> – temperature of the surrounding.<\/p>\n

            Question 7.
            \nState and explain Stefan\u2019s law.
            \nAnswer:
            \nThe total radiation emitted by a perfect black body per second per unit area of the surface is directly proportional to the fourth power of its absolute temperature. If E is radiation emitted per second per unit area of a perfect black body at temperature T then, E \u221d T4<\/sup> or E = \u03c3T4<\/sup> where s is called Stefan\u2019s constant.<\/p>\n

            Question 8.
            \nState and explain Wein\u2019s displacement law.
            \nAnswer:
            \nThe wavelength (\u03bbm<\/sub>) corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature.
            \ni. e., \u03bbm<\/sub> \u221d \\(\\frac{1}{T}\\) or \u03bbm<\/sub> T = constant.
            \nThe constant is called Wein\u2019s Constant.<\/p>\n

            Question 9.
            \nState and explain Planck\u2019s law.
            \nAnswer:
            \nAccording to this law, radiation of a particular frequency (v) is emitted or absorbed in discrete amounts (E) given by the relation E = h v where h is a Planck\u2019s constant. The value of h is 6.625 \u00d7 10-34<\/sup> Js. These packets of energy are called quantum or photon.<\/p>\n

            \"KSEEB<\/p>\n

            Question 10.
            \nDefine solar constant. What is its value?
            \nAnswer:
            \nThe solar constant is defined as the amount of heat radiation absorbed normally per second per unit area of a perfect black body placed on the earth at a mean distance from the sun in the absence of atmosphere.
            \nThe value of solar Constant is 1350 W\/m2<\/sup>.<\/p>\n

            Question 11.
            \nState<\/p>\n

              \n
            1. Kirchhoff\u2019s law of radiation,<\/li>\n
            2. State Stefan\u2019s law of radiation<\/li>\n<\/ol>\n

              Answer:
              \n1. Kirchhoff\u2019s law:
              \nAt a given temperature, the ratio of the emissive power to the absorptive power is a constant for all bodies and is equal to the emissive power of a perfect black body.<\/p>\n

              2. Stefan\u2019s law:
              \nThe total radiation emitted by a perfectly black body per second per unit area is directly proportional to the 4th<\/sup> power of its absolute temperature.<\/p>\n

              Question 12.
              \nCalculate the heat of combustion of coal, when 10 gm of coal, on burning raises the temperature of 2 litres of water from 20\u00b0 C to 55\u00b0 C.
              \nAnswer:
              \nmass of 2l of water = 2kg
              \n\u2206T = 55\u00b0 C – 20\u00b0 C = 35\u00b0 C
              \ns = 4.2 \u00d7 103<\/sup> J kg-1<\/sup> \u00b0C-1<\/sup>
              \nHeat produced by burning coal = ms \u2206T
              \n= 2 \u00d7 4.2 \u00d7 103<\/sup> \u00d7 35
              \n= 2.94 \u00d7 105<\/sup> J<\/p>\n

              \"KSEEB<\/p>\n

              Question 13.
              \n2 kg of water at 80\u00b0 C is mixed with 3 kg of water at 20\u00b0C. Assuming no heat losses, find the final temperature of the mixture.
              \nAnswer:
              \nm1<\/sub> = 2 kg,
              \nT1<\/sub> = 80\u00b0C,
              \ns = 4.2 \u00d7 103<\/sup> J kg-1<\/sup> \u00b0C-1<\/sup>,
              \nm2<\/sub> = 3kg,
              \nT2<\/sub> = 20\u00b0 C
              \nLet the final temperature be T3<\/sub>
              \nHeat lost by 3 kg water = Heat gained by 2 kg water.
              \n\u21d2 m1<\/sub>s(T1<\/sub> – T3<\/sub>) = m2<\/sub>s(T3<\/sub> – T2<\/sub>)
              \n\u21d2 2 \u00d7 (80 – T3<\/sub>) = 3 \u00d7 (T3<\/sub> – 20)
              \n\u21d2 T3<\/sub> = 44\u00b0 C
              \n\u2234 Final temperature = 44\u00b0 C<\/p>\n

              Question 14.
              \nWhat are the basic requirements of a cooking utensil in respect of:<\/p>\n

                \n
              1. Specific heat and<\/li>\n
              2. thermal conductivity<\/li>\n<\/ol>\n

                Answer:
                \nFor the temperature of the utensil to rise quickly, it is necessary for the utensil to have low specific heat. In order for the heat to spread quickly to the vegetables, the utensil must have high thermal conductivity.<\/p>\n

                1st PUC Physics Thermal Properties of Matter Three Marks Questions and Answers<\/h3>\n

                Question 1.
                \nState Wien\u2019s displacement law. Draw the graph showing energy emitted versus wavelength for a black body at different temperatures.
                \nAnswer:
                \nWein\u2019s displacement law:
                \nThe wavelength (\u03bbm<\/sub>) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. \u03bbm<\/sub> \u221d \\(\\frac{1}{T}\\) or \u03bbm<\/sub> T = b. where b is a constant, called Wein\u2019s constant. The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures is as shown in the figure.
                \n\"1st
                \nThe energy distribution is not uniform. There is a particular wavelength \u03bbm<\/sub> at which the energy emitted is maximum. The wavelength \u03bbm<\/sub> for which the intensity is maximum decreases with increase in temperature.<\/p>\n

                Question 2.
                \nDraw energy distribution curves for a black body at two different temperatures T1<\/sub> and T2<\/sub> (T1<\/sub> > T2<\/sub>). Write any two conclusions that can be drawn from these curves.
                \nAnswer:
                \nThe radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength is as shown in the figure.
                \n\"1st<\/p>\n

                \"KSEEB<\/p>\n

                Question 3.
                \nDraw labelled curves which show how the energy in the black body radiation spectrum varies with wavelength at various temperatures. What result does this lead to?
                \nAnswer:
                \nThe radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength is as shown in the figure.
                \n\"1st<\/p>\n

                1st PUC Physics Thermal Properties of Matter Four Or Five Marks Questions and Answers<\/h3>\n

                Question 1.
                \nState and explain the laws of thermal conductivity.
                \nAnswer:
                \n\"1st
                \nConsider a metal slab of area of cross-section A and thickness d. Let the end faces (ABCD & EFGH) be maintained at temperatures \u03b81<\/sub> and \u03b82<\/sub> (\u03b81<\/sub> > \u03b82<\/sub>).
                \nThe amount of heat conducted (Q) from the hotter to colder face is,
                \n1. directly proportional to the area of cross-section (A)
                \n2. directly proportional to the time for which heat flows (t)
                \n3. directly proportional to the temperature difference between the faces (\u03b81<\/sub> – \u03b82<\/sub>)
                \n4. inversely proportional to the distance between the faces (d)
                \nThis is called the law of thermal conductivity.
                \nAccording to the law Q \u221d \\(\\frac{\\mathrm{At}\\left(\\theta_{1}-\\theta_{2}\\right)}{\\mathrm{d}}\\)
                \n\\(\\mathrm{Q}=\\frac{\\mathrm{KAt}\\left(\\theta_{1}-\\theta_{2}\\right)}{\\mathrm{d}}\\)
                \nK is a constant of proportionality called coefficient of thermal conductivity.<\/p>\n

                Question 2.
                \nWhat is black body radiation? Explain the characteristics of It.
                \nAnswer:
                \nThe radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength is as shown in the figure.
                \n\"1st
                \nCharacteristics of black body radiation<\/p>\n

                  \n
                1. A black body emits radiation of all wavelengths lying in the region of ultraviolet, visible light and infrared.<\/li>\n
                2. As the temperature of the black body increases, the total amount of radiation emitted by it increases. (According to Stefan\u2019s law)<\/li>\n
                3. The energy distribution is not uniform. There is a particular wavelength \u03bbm<\/sub> at which the energy emitted is maximum.<\/li>\n
                4. The wavelength \u03bbm<\/sub> for which the energy emitted is maximum decreases with increase in temperature.<\/li>\n<\/ol>\n

                  \"KSEEB<\/p>\n

                  Question 3.
                  \nDescribe a total radiation pyrometer.
                  \nAnswer:
                  \n\"1st
                  \nIt consists of a concave mirror M made of copper and plated with nickel. The mirror has a small hole at the centre at which an eyepiece E is fitted. A diaphragm D is placed at the focus of M; which has a hole to allow radiation to pass through it. A blackened metal strip S is placed behind D. To the other surface of S, one junction of a thermocouple TT is connected. The thermocouple is connected to a millivoltmeter. A screen R is used to protect the strip from direct radiation. The mirror can be adjusted for focusing.<\/p>\n

                  Question 4.
                  \nWhy is ice packed in sawdust? Explain.
                  \nAnswer:
                  \nIce is packed in sawdust because it prevents ice from melting. Sawdust is a bad conductor of heat.<\/p>\n

                  Question 5.
                  \nExplain the different modes of transmission of heat. Explain Wein\u2019s displacement law with the help of graph of intensity versus wavelength.
                  \nAnswer:
                  \n1. Conduction:
                  \nIt is a process of transmission of heat in solids, without the actual movement of the particles.
                  \nEg: When metal rod is heated at one end, after some time another end also becomes hot.<\/p>\n

                  2. Convection:
                  \nIt is a process of transmission of heat in fluids (liquids and gases) with the actual movement of particles.
                  \nEg: Smoke carries heat energy from the flame by convection.<\/p>\n

                  3. Radiation:
                  \nIt is a process of transmission of heat without any medium in the form an electromagnetic waves.
                  \nEg: Heat from the sun reaches the earth by radiation.<\/p>\n

                  Wein\u2019s displacement law:
                  \nThe wavelength (\u03bbm<\/sub>) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. \u03bbm<\/sub> \u221d \\(\\frac{1}{T}\\) or \u03bbm<\/sub> T = b. where b is a constant, called Wein\u2019s constant. The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures is as shown in the figure.
                  \n\"1st
                  \nThe energy distribution is not uniform. There is a particular wavelength \u03bbm<\/sub> at which the energy emitted is maximum. The wavelength \u03bbm<\/sub> for which the intensity is maximum decreases with increase in temperature.<\/p>\n

                  Question 6.
                  \nWhat is a black body? Draw the curves showing the energy distribution among black body radiations at different temperature. Hence define Wein’s displacement law. Give one application of Wein’s displacement law.
                  \nAnswer:
                  \n1. black body:
                  \nThe radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength.
                  \n\"1st
                  \nWein\u2019s displacement law:
                  \nThe wavelength (\u03bbm<\/sub>) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. \u03bbm<\/sub> \u221d \\(\\frac{1}{T}\\) 17 or \u03bbm<\/sub> T = b. where b is a constant, called Wein\u2019s constant.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 7.
                  \nDiscuss briefly energy distribution of black body radiation. Hence deduce Wien\u2019s displacement law and Stefan\u2019s law.
                  \nAnswer:
                  \nWein\u2019s displacement law:
                  \nThe wavelength (\u03bbm<\/sub>) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. \u03bbm<\/sub> \u221d \\(\\frac{1}{T}\\) or \u03bbm<\/sub> T = b. where b is a constant, called Wein\u2019s constant. The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures is as shown in the figure.
                  \n\"1st
                  \nThe energy distribution is not uniform. There is a particular wavelength \u03bbm<\/sub> at which the energy emitted is maximum. The wavelength \u03bbm<\/sub> for which the intensity is maximum decreases with increase in temperature.<\/p>\n

                  Stefan\u2019s law:
                  \nThe total radiation emitted by a perfect black body per second per unit area of the surface is directly proportional to the fourth power of its absolute temperature. If E is radiation emitted per second per unit area of a perfect black body at temperature T then, E \u221d T4<\/sup> or E = \u03c3T4<\/sup> where s is called Stefan\u2019s constant.<\/p>\n

                  1st PUC Physics Thermal Properties of Matter Numerical Problems Questions and Answers<\/strong><\/p>\n

                  Question 1.
                  \nA circular plate has a radius of 10 cm and thickness 1 cm. If one face of the plate is kept in contact with steam and the other face is in contact with ice, calculate the amount of heat flowing per hour across the plate. K as 85 W-1<\/sup>m-1<\/sup>K-1<\/sup>.
                  \nSolution:
                  \nThe amount of heat conducted in t seconds is given by,
                  \n\\(Q=K \\frac{A\\left(\\theta_{1}-\\theta_{2}\\right) t}{d}\\)
                  \nHere, K = 85 W m-1<\/sup>K-1<\/sup>
                  \nA = \u03c0 r2<\/sup> = 3.14 \u00d7 (0.1)2<\/sup>
                  \nt = 1 hr = 60 \u00d7 60 s.
                  \nd = 10-2<\/sup> m2<\/sup>
                  \n(\u03b81<\/sub> – \u03b82<\/sub>) = 373 -273 = 100 K
                  \nQ= \\(\\mathrm{K} \\frac{\\mathrm{A}\\left(\\theta_{1}-\\theta_{2}\\right) \\mathrm{t}}{\\mathrm{d}}\\)
                  \n= \\(\\frac{85 \\times 3.14 \\times(0.1)^{2} \\times 100 \\times 60 \\times 60}{10^{-2}}\\)
                  \n= 96.08×106<\/sup> J.<\/p>\n

                  Question 2.
                  \nA metal rod 0.314 m long and 0.02 m in diameter has one end in boiling water at 373K and the other end in melting ice at 273K. Find the amount of ice that will melt per hour? K = 385 Wm-1<\/sup>K-1<\/sup>.
                  \nLatent heat of Ice = 3.36 \u00d7 105<\/sup> Jkg-1<\/sup>.
                  \nSolution:
                  \nIf \u2018m\u2019 is mass of ice that melts in one hour, then quantity of heat conducted is Q = mL
                  \nBut Q = K \\(\\frac{A\\left(\\theta_{1}-\\theta_{2}\\right) t}{d}\\)
                  \n\u2234 Q = mL
                  \nm = \\(\\frac{\\mathrm{KA}\\left(\\theta_{1}-\\theta_{2}\\right) t}{\\mathrm{d} \\times L}\\)
                  \nHere, K =385Wm-1<\/sup>K-1<\/sup>
                  \nA = p r2<\/sup> =3.14 \u00d7 (0.01 )2<\/sup>
                  \nL = 3.3 \u00d7 105<\/sup> Jkg-1<\/sup>
                  \n\u03b81<\/sub> – \u03b82<\/sub> = 373 – 273 = 100 K
                  \nt = 1 hour = 3600 s
                  \nd = 0.314 m
                  \n\u2234 m= \\(\\frac{385 \\times 3.142 \\times(0.01)^{2} \\times 100 \\times 60 \\times 60}{0.314 \\times 3.36 \\times 10^{5}}\\)
                  \n= 41.28 x 10-2<\/sup>kg<\/p>\n

                  Question 3.
                  \nTwo metal rods of the same area of cross-section and with their lengths in the ratio 1:3 are joined together to form a compound rod. One end of this compound rod is kept at 100\u00b0C and the other at 0\u00b0 C. Calculate the temperature at the Junction if the thermal conductivity is in the ratio 1:3.
                  \nSolution:
                  \n\"1st
                  \nLet AB and BC be the two rods of length L1<\/sub> and L2<\/sub> and having the same area of cross-section A, joined together to form a compound rod as shown in the figure.
                  \nIf \u2018 \u03b8 \u2019 is the steady temperature at the junction B during the steady-state, then, rate of flow of heat through AB = rate of flow through BC
                  \n\"1st<\/p>\n

                  Question 4.
                  \nCalculate rate of heat radiated per unit area from the surface of a furnace at a temperature 2000K, assuming it to be a black body. If the emissivity is 0.37, then what Is the amount of heat radiated per second? G =5.67×10-8<\/sup>Wm-2<\/sup>K-4<\/sup>.
                  \nSolution:
                  \nFrom Stefan\u2019s law of radiation, the amount of heat radiated per second is E = \u03c3T4<\/sup>.
                  \nE =5.67 \u00d7 10-8<\/sup> \u00d7 (2000)4<\/sup>
                  \n= 5.67 \u00d7 24<\/sup> \u00d7 104<\/sup>
                  \n= 5.67 \u00d7 16 \u00d7 104<\/sup>
                  \n= 9.072 \u00d7 105<\/sup>Js-1<\/sup>.
                  \nIf e is the emissivity of the surface, then the heat radiated per second R1<\/sup> = eR
                  \n= 0.37 \u00d7 9.072 \u00d7 105<\/sup>
                  \n= 3.35 \u00d7 105<\/sup> Js-1<\/sup><\/p>\n

                  \"KSEEB<\/p>\n

                  Question 5.
                  \nA slab of stone of area 0.36m2<\/sup> and thickness 0.1m is exposed on the lower surface to steam at 373 K. A block of ice at 273 k rests on the upper surface of the slab. If in one hour, 4.8kg of Ice is melted, calculate the conductivity of the stone. Latent heat of Ice = 3.35 x 105<\/sup> JK-1<\/sup>.
                  \nSolution:
                  \nA = 0.36m2<\/sup>,
                  \nd = 0.1m
                  \n\u03b81<\/sub> = 373 K
                  \n\u03b82<\/sub> = 273 K
                  \nt = 1 hr = 3600s,
                  \nm = 4.8kg,
                  \nK = ?
                  \nL = 3.35 \u00d7 105<\/sup> Jkg-1<\/sup>
                  \nHeat conducted through the rod = Heal used in melting ice
                  \n\\(\\frac{\\mathrm{KA}\\left(\\theta_{1}-\\theta_{2}\\right) \\mathrm{t}}{\\mathrm{d}}\\) = mL
                  \nK = \\(\\frac{m L d}{A\\left(\\theta_{1}-\\theta_{2}\\right) t}=\\frac{4.8 \\times 3.35 \\times 10^{5} \\times 0.1}{0.36 \\times 100 \\times 3600}\\)
                  \n= 1.241 Wm-1<\/sup> K-1<\/sup>.<\/p>\n

                  Question 6.
                  \nThe filament of 40 W electric lamp of length 0.1m and diameter 0.1mm is kept at a temperature of 2773 K. Calculate the emissive power if all the heat is lost by radiation. \u03c3 = 5.67 \u00d7 10-8<\/sup> Wm-2<\/sup> K-4<\/sup>].
                  \nSolution :
                  \nP = 40 W,
                  \nl = 0.1 m,
                  \nr = 0.05 mm
                  \nT = 2773 K
                  \ne = ?
                  \nArea = 2\u03c0 rl =3.142 x 10-5<\/sup> m2<\/sup>
                  \nEnergy emitted by a source in t sec
                  \nE = e A \u03c3 T4<\/sup>t
                  \n\\(\\frac{E}{t}\\) = e A \u03c3 T4<\/sup>
                  \nP = e A \u03c3 T4<\/sup>
                  \ne = \\(\\frac{P}{A \\sigma T^{4}}\\)
                  \n\\(=\\frac{40}{3.142 \\times 10^{-5} \\times 5.67 \\times 10^{-8} \\times(2773)^{4}}\\)
                  \n= 0.3798<\/p>\n

                  Question 7.
                  \nThe thickness of ice in a lake at a certain moment is 0.03m. At what rate the thickness of the ice increasing and how long will it take for the thickness of ice layer to be doubled, if the temperature of air is -20\u00b0 C. Given coefficient of thermal conductivity of ice=0.168 Wm-1<\/sup>K-1<\/sup> degree. Latent heat of fusion of ice=3.35 \u00d7 10 Jkg-1<\/sup> density of ice= 9.2 \u00d7 102<\/sup> Kg\/m3<\/sup>
                  \nSolution:
                  \nWhen the layer of ice covers the lake, the lower part of the ice is in contact with ice is at 0\u00b0c and outer surface with air at – 20\u00b0c
                  \n\u2234 \u03b81<\/sub> =0\u00b0c and \u03b82<\/sub> = – 20\u00b0c
                  \nLet the thickness of ice increase by small amount \u2206x by an area A in a time \u2206t.
                  \nThen,
                  \nmass of ice formed = A \u00d7 \u2206x \u00d7 \u03c1
                  \nQuantity of heat, Q = A \u00d7 \u2206x \u00d7 \u03c1 \u00d7 L\u00a0 \u00a0 …..(1)
                  \nBut, Q = \\(\\frac{\\mathrm{k} \\mathrm{A}\\left(\\theta_{1}-\\theta_{2}\\right) \\Delta \\mathrm{t}}{\\mathrm{d}}\\) ….(2)
                  \nfrom equation (1) & equation (2)
                  \nA \u00d7 \u2206x \u00d7 r \u00d7 L = \\(\\frac{\\mathrm{k} \\mathrm{A}\\left(\\theta_{1}-\\theta_{2}\\right) \\Delta \\mathrm{t}}{\\mathrm{d}}\\)
                  \n\\(\\frac{\\Delta x}{\\Delta t}=\\frac{k A\\left(\\theta_{1}-\\theta_{2}\\right)}{d x A x \\rho x L}\\)
                  \n\\(=\\frac{0.168 \\times 20}{0.03 \\times 9.2 \\times 10^{2} \\times 3.35 \\times 10^{5}}\\)
                  \n= 3.634 \u00d7 10-7<\/sup>ms-1<\/sup>
                  \nThe time taken for the ice layer to double is
                  \nt = \\(\\frac{\\Delta x \/ \\Delta t}{d}\\)
                  \n\\(=\\frac{3.634 \\times 10^{-7}}{0.06}\\)
                  \n= 60.57 \u00d7 10-7<\/sup>sec<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 8.
                  \nEstimate the rate at which ice melts in a wooden box 0.04m thick, of inside measurement 0.6 m \u00d7 0.4m \u00d7 0.4m. Assume that the outside temperature is 300 K and the conductivity of wood is 0.08 wm-1<\/sup>K-1<\/sup>
                  \nSolution:
                  \nArea of the box
                  \n= A = 4 \u00d7 (0.6 \u00d7 0.4) + 2 \u00d7 (0.4 \u00d7 0.4)= 1.28m2<\/sup>
                  \nd = 0.04m
                  \nk= 0.08wm-1<\/sup>K-1<\/sup>
                  \nL= 3.36 \u00d7 105<\/sup>Jkg-1<\/sup>
                  \nLet Q be the heat transmitted per second.
                  \nThen, Q = \\(\\frac{\\mathrm{k} \\mathrm{A}\\left(\\theta_{1}-\\theta_{2}\\right) \\mathrm{t}}{\\mathrm{d}}\\)
                  \n\\(=\\frac{0.08 \\times 1.28 \\times(300-273) \\times 1}{0.04}=69.12 J\\)
                  \nTo melt 1 kg of ice, 3.36 \u00d7 105<\/sup>J of heat is required
                  \n\u2234 Rate at which ice melts
                  \n\\(=\\frac{69.12}{3.36 \\times 10^{-5}}\\)
                  \n= 20.57 \u00d7 10-5<\/sup>kgs-1<\/sup>.<\/p>\n

                  Question 9.
                  \nA man wraps himself in flannel 4 mm thick. How much heat will be lost per square metre of his body per hour if the atmosphere temperature is 20\u00b0 and the body temperature is 37\u00b0c(The co-efficient of thermal conductivity of flannel is 0.05 Wm-1<\/sup>k-1<\/sup>)
                  \nSolution:
                  \nk=0.05Wm-1<\/sup>k-1<\/sup>
                  \nd = 4mm = 4 \u00d7 103<\/sup>
                  \nm = 0.004m
                  \n\u03b81<\/sub> =37\u00b0c,
                  \n\u03b82<\/sub> =20\u00b0c
                  \nt =1 hr=3600 seconds
                  \nA =1m2<\/sup>
                  \nThe quantity of heat conducted is
                  \n\\(Q=\\frac{k A\\left(\\theta_{1}-\\theta_{2}\\right) t}{d}\\)
                  \n\\(=\\frac{0.05 \\times 1(37-20) 3600}{0.004}\\)
                  \n= 7.65 \u00d7 105<\/sup>J<\/p>\n

                  Question 10.
                  \nCalculate the rate at which heat is lost through a glass window of area 0.1m2<\/sup> thickness 4mm when the temperature inside is at 40\u00b0 c and outside is -5\u00b0c. Co-efficient of thermal conductivity of a glass is 1.1wm-1<\/sup>k-1<\/sup>.
                  \nSolution:
                  \nA=0.1m2<\/sup>
                  \nd = 4mm = 4 \u00d7 10-3<\/sup>
                  \nm=0.004m
                  \n\u03b81<\/sub> = 40\u00b0C
                  \n\u03b82<\/sub> = -5\u00b0c
                  \nk = 1.1Wm-1<\/sup>k-1<\/sup>
                  \nHeat lost through the glass window is
                  \n\\(Q=\\frac{K A\\left(\\theta_{1}-\\theta_{2}\\right) t}{d}\\)
                  \n\\(=\\frac{1.1 \\times 0.1(40+5) \\times 1}{0.004}\\)
                  \n= 1237.5J<\/p>\n

                  Question 11.
                  \nThe two faces of metal plate of thickness 40mm and area 102<\/sup>m2<\/sup> maintained at 273K and 373K. The coefficient of thermal conductivity of the metal is 380 Wm-1<\/sup>k-1<\/sup> Find the quantity of heat flows, through the plate in one hour.
                  \nSolution:
                  \nd = 40mm = 40 \u00d7 10-3<\/sup>
                  \nm = 0.04m
                  \nA = 102<\/sup>m2<\/sup>
                  \n\u03b82<\/sub> = 273k
                  \n\u03b81<\/sub> =37K
                  \nK = 380Wm-1<\/sup>k-1<\/sup>
                  \nt = 1hr = 3600 seconds
                  \nThe quantity of heat that flows in one hour is
                  \n\\(Q=\\frac{k A\\left(\\theta_{1}-\\theta_{2}\\right) t}{d}\\)
                  \n\\(=\\frac{380 \\times 10^{2}(373-273) \\times 3600}{0.04}\\)
                  \nQ = 3.42 \u00d7 1011<\/sup>J<\/p>\n

                  Question 12.
                  \nA spherical body of radius 0.01m is at 400K. Calculate the energy radiated by the body per second per unit area. Assume the body to be perfect black body.
                  \nSolution:
                  \nFrom stefan\u2019s law of radiation, amount of heat radiated per second per unit area is
                  \nE = \u03c3 T4<\/sup>
                  \n= 5.67 \u00d7 10-8<\/sup> \u00d7 (400)4<\/sup> = 1451.5s-1<\/sup>
                  \nThe total energy radiated per second is
                  \nE = A \u03c3 T4<\/sup>
                  \nE = P r2<\/sup> \u03c3 T4<\/sup>
                  \n= 3.142 \u00d7 (0.01 )2<\/sup> \u00d7 5.67 \u00d7 10-8<\/sup> \u00d7 (400)4<\/sup>
                  \n= 0.4561s-1<\/sup><\/p>\n

                  Question 13.
                  \nA metal rod 0.3m long & 0.02m in diameter has one end at 100\u00b0C & other end at 0\u00b0C. Calculate the total amount of heat conducted in 2 minutes. Given k = 385 JS-1<\/sup>m-1<\/sup>K-1<\/sup>.
                  \nSolution:
                  \nl = 0.3m,
                  \nd = 0.02m
                  \n\u2234 r = 0.01m
                  \n\u03b81<\/sub> = 100\u00b0C
                  \n\u03b82<\/sub> = O\u00b0C
                  \nt = 2 minutes = 120 S
                  \nk = 385 JS-1<\/sup>m-1<\/sup>K-1<\/sup>.
                  \nQ = ?
                  \nt= \u03c0 r2<\/sup> = 3.142 \u00d7 0.012<\/sup> = 3.142 \u00d7 104<\/sup>m2<\/sup>
                  \n\\(\\therefore Q=\\frac{k A t\\left(\\theta_{1}-\\theta_{2}\\right)}{\\ell}\\)
                  \n\\(=\\frac{385 \\times 3.142 \\times 10^{4} \\times 120(109)}{0.3}\\)
                  \n= 4.8 \u00d7 103<\/sup> J<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 14.
                  \nA metallic rod 1.5m long and 2×104<\/sup> m2<\/sup> in area is heated at one end. The coefficient of thermal conductivity of the material of the metallic rod is 210 Js-1<\/sup>m-1<\/sup>K-1<\/sup>. In the steady-state the temperature of the ends of the rod are 373K and 263 K. Calculate.<\/p>\n

                    \n
                  1. Temperature gradient In the rod.<\/li>\n
                  2. Temperature at a point 0.8 m from the hot end.<\/li>\n
                  3. Rate of heat transmission.<\/li>\n<\/ol>\n

                    Answer:
                    \nd=1.5m,
                    \nA = 2 \u00d7 104<\/sup>m2<\/sup>,
                    \nK = 210 W\/m\/k, \u03b81<\/sub> = 373K and \u03b82<\/sub> = 263K.
                    \n1. Temperature gradient
                    \n\\(\\frac{\\theta_{1}-\\theta_{2}}{d}=\\frac{373-263}{1.5}=\\frac{110}{1.5}=73.33 \\mathrm{K} \/ \\mathrm{m}\\)<\/p>\n

                    2. Let \u03b8 be the temperature at a distance d1<\/sup> = 0.8m from the hot end then temperature gradient = \\(\\frac{\\theta_{1}-\\theta}{d^{1}}\\)
                    \ni.e., 73.33 = \\(\\frac{373-\\theta}{0.8}\\) or \u03b8 =314.336 K.<\/p>\n

                    3. Rate of heat transmission
                    \n\\(\\frac{\\mathrm{q}}{\\mathrm{t}}=\\mathrm{KA}\\left(\\frac{\\theta_{1}-\\theta_{2}}{\\mathrm{d}}\\right)\\)
                    \n= 210 \u00d7 2 \u00d7 10-4<\/sup> \u00d7 73.33
                    \n= 3.08 J\/s.<\/p>\n

                    Question 15.
                    \nThe two faces of a metal plate of thickness 40 mm and area 10-2<\/sup>m2<\/sup> are maintained at 273 K and 373 K. The coefficient of thermal conductivity of metal is 380 w\/m\/K. Find the quantity of heat that flows through the plate in one hour.
                    \nAnswer:
                    \nd = 40 \u00d7 10-3<\/sup>m,
                    \nA = 10-2<\/sup>m2<\/sup>,
                    \nt = 1 hr = 3600s,
                    \n\u03b81<\/sub> = 373K,
                    \n\u03b81<\/sub> = 2.73K and K= 380 W\/m\/K
                    \n\\(\\mathrm{q}=\\frac{\\kappa \\mathrm{At}\\left(\\mathrm{q}-\\theta_{2}\\right)}{\\mathrm{d}}\\)
                    \n= 34.6×106<\/sup>J.<\/p>\n

                    Question 16.
                    \nThe time taken for a substance to cool form 100\u00b0 C to 50\u00b0 C is 5 minutes. If the room temperature is 30\u00b0C, what time required for same substance to cool from 50\u00b0C to 40\u00b0C?
                    \nAnswer:
                    \nCase (i):
                    \nRate of cooling
                    \n\\(\\frac{\\mathrm{d} \\theta}{\\mathrm{dt}}=\\frac{100-50}{5}\\) = 10\u00b0 C\/min,
                    \nAverage temperature
                    \n\\(\\theta=\\frac{100+50}{2}=75^{\\circ} \\mathrm{C}\\)
                    \nRoom temperate \u03b80<\/sub> = 30\u00b0C.
                    \nAccording to Newton\u2019s law of cooling,
                    \n\\(\\frac{\\mathrm{d} \\theta}{\\mathrm{dt}}\\) = -K(\u03b8 – \u03b80<\/sub>).
                    \n10 = -K(75 – 30).
                    \nK= \\(-\\frac{10}{45} \/ \\mathrm{min}\\) . ……(1).<\/p>\n

                    Case(ii):
                    \nRate of cooling
                    \n\\(\\frac{\\mathrm{d} \\theta}{\\mathrm{dt}}\\) = \\(\\frac{50-40}{\\mathrm{dt}}=\\frac{10}{\\mathrm{dt}}\\),
                    \nAverage temperature
                    \n\\(\\theta=\\frac{50+40}{2}=45^{\\circ} \\mathrm{C}\\)
                    \nRoom temperate e \u03b80<\/sub> = 30\u00b0C.
                    \nAccording to Newton\u2019s law of cooling,
                    \n\\(\\frac{\\mathrm{d} \\theta}{\\mathrm{dt}}\\) = K(\u03b8 – \u03b80<\/sub>)
                    \n\\(\\frac{10}{\\mathrm{dt}}=-\\left(\\frac{-10}{45}\\right)(45-30)\\) (from)
                    \ndt =3 min.<\/p>\n

                    1st PUC Physics Thermal Properties of Matter Hard Questions and Answers<\/strong><\/p>\n

                    Question 1.
                    \nA hot body placed in air is cooled down. According to Newton\u2019s law of cooling, the rate of decrease of temperature being \u2018k\u2019 times the temperature of the surrounding. Starting from t = 0, find the time in which the body will lose 90% of the maximum heat it can lose.
                    \nAnswer:
                    \nWe have,
                    \n\\(\\frac{\\mathrm{d} \\mathrm{T}}{\\mathrm{dt}}\\) = – K(T – T0<\/sub>)
                    \nwhere T0<\/sub> is the temperature of the surrounding and T is the temperature of the body at time t. Let T = T1<\/sub> at t = 0.
                    \nThen
                    \n\"1st
                    \n\u21d2 T – T0<\/sub> = (T1<\/sub> – T0<\/sub>) e-kt<\/sup> ……(1)
                    \nThe body continues to lose heat until its temperature becomes equal to that of the surrounding. The loss of heat in this entire period is
                    \n\u2206H = ms(T -T0<\/sub>).
                    \nThis is the maximum heat it can lose. If the body loses 90% of this heat, the decrease in to temperature will be
                    \n\\(\\frac{9 \\Delta \\mathrm{H}}{10 \\mathrm{ms}}\\) = \\(\\frac{9}{10}\\) (T1<\/sub> – T0<\/sub>)
                    \nIf the body loses this heat in time t, the temperature at t1<\/sub> will be
                    \nT1<\/sub> – \\(\\frac{9}{10}\\) (T1<\/sub> – T0<\/sub>)
                    \n\\(=\\frac{T_{1}}{10}+\\frac{9 T_{0}}{10}\\)
                    \nPutting these values of time and temperature in (1),
                    \n\\(\\frac{T_{1}+9 T_{0}}{10}\\) – T0<\/sub> = (T1<\/sub> – T2<\/sub>) e-kt1<\/sub><\/sup>
                    \n\u21d2 \\(\\frac{T_{1}-T_{0}}{10}\\) = (T1<\/sub> – T2<\/sub>) e-kt1<\/sub><\/sup>
                    \n\u21d2 e-kt1<\/sub><\/sup> = 1\/10
                    \n\u21d2 t1<\/sub> = \\(\\frac{\\ln 10}{k}\\)
                    \n\u2234 Required time = \\(\\frac{\\ln 10}{k}\\)<\/p>\n

                    \"KSEEB<\/p>\n

                    Question 2.
                    \nTwo thin metallic spherical shells of radii r1<\/sub> and r2<\/sub> (r1<\/sub> < r2<\/sub> are placed with their centres coinciding. A material of thermal conductivity k is filled in the space between the shells. The inner shell is maintained at temperature T1<\/sub> and the outer shell at temperature T2<\/sub> (T1<\/sub> < T2<\/sub>). Calculate the rate at which heat flows radially through the material.
                    \nAnswer:
                    \n\"1st
                    \nConsider two spherical shells of radii x and x + dx concentric with the given system. Let the temperatures at these shells be T and T + dt respectively. The amount of heat flowing radially inward through the material between x and x + dx is
                    \n\"1st<\/p>\n","protected":false},"excerpt":{"rendered":"

                    You can Download Chapter 11 Thermal Properties of Matter Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Physics Question Bank Chapter 11 Thermal Properties of Matter 1st PUC Physics Thermal Properties of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9999"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=9999"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9999\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9999"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9999"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9999"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}