CBSE Class 12 Physics Heat and Thermodynamics Solved Examples

CBSE Class 12 Physics Heat and Thermodynamics Solved Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

SOLVED EXAMPLES
CLASS XII : PHYSICS
HEAT AND THERMODYNAMICS

Ex.1 The temperatures of equal masses of three different liquids A, B and C are 12ºC, 19ºC and 28ºC respectively. The temperature when A and B are mixed is 16ºC and when B and C are mixed, it is 23ºC. Find the temperature of mixture when A and C liquids are mixed.

Answer:
1. Let m be the mass of each liquid and S_A, S_B, S_C their specific heats.
2. A and B mixed at 16ºC: heat lost by B = heat gained by A, so mS_B(19 - 16) = mS_A(16 - 12), which gives 3S_B = 4S_A.
3. B and C mixed at 23ºC: mS_C(28 - 23) = mS_B(23 - 19), which gives 5S_C = 4S_B.
4. From steps 2 and 3, S_B = \( \frac{4}{3} \) S_A and S_C = \( \frac{4}{5} \) S_B = \( \frac{16}{15} \) S_A.
5. Let T be the temperature when A and C are mixed: mS_C(28 - T) = mS_A(T - 12).
6. \( \frac{16}{15} \)(28 - T) = T - 12, so 448 - 16T = 15T - 180, which gives 31T = 628.
7. T = \( \frac{628}{31} \) ≈ 20.2ºC.

Teacher's Note:
a) Equal masses are given, so the mass cancels in every heat balance.
b) The value 628/31 gives 20.2ºC.
c) Reduce each specific heat to S_A first, then the final equation is easy to solve.

Ex.2 A mixture of 250 gm of water and 200 gm of ice at 0ºC is kept in a calorimeter which has a water equivalent of 50 gm. If 200 gm of steam at 100ºC is passed through this mixture, calculate the final temperature and weight of the contents of the calorimeter. Latent heat of fusion of ice = 80 cal/gm., Latent heat of vaporization of water to steam = 540 cal/gm., Specific heat of water = 1 cal/gm/ºC.

Answer:
1. Heat given out by 200 gm of steam when it condenses at 100ºC: 200 × 540 = 108000 cal.
2. Heat needed by 200 gm of ice at 0ºC to melt and warm to 100ºC: 200 × 80 + 200 × 1 × (100 - 0) = 36000 cal.
3. Heat needed by 250 gm of water and the calorimeter (water equivalent 50 gm) to warm from 0ºC to 100ºC: 300 × 1 × (100 - 0) = 30000 cal.
4. Total heat needed = 36000 + 30000 = 66000 cal, which is less than 108000 cal, so only part of the steam condenses.
5. Mass of steam condensed M: M × 540 = 66000, so M = \( \frac{66000}{540} \) = 122.2 gm.
6. Final temperature of the system = 100ºC.
7. Weight of the contents = 250 + 200 + 122.2 = 572.2 gm.
8. Hence, the final temperature is 100ºC and the contents weigh 572.2 gm.

Teacher's Note:
a) Compare the total heat available from the steam with the heat needed by the ice, water and calorimeter, to find how much steam condenses.
b) Since some steam remains, the final temperature is 100ºC, the boiling point of water.
c) The calorimeter itself is not counted in the weight of contents, which is why 50 gm appears only as water equivalent.

Ex.3 A copper calorimeter of mass 100 gm contains 200 g of a mixture of ice and water. Steam at 100ºC under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to 50ºC. If the mass of the calorimeter and its contents is now 330 gm, what was the ratio of ice and water in the beginning ? Neglect heat losses. Given that :
Specific heat of copper = 0.42 × 10³ J/kgº Kº
Specific heat of water = 4.2 × 10³ J/kg Kº
Latent heat of fusion of ice = 3.36 × 10⁵ J/kg Kº
Latent heat of condensation of steam = 22.5 × 10⁵ J/kg K.

Answer:
1. Let the calorimeter contain originally x gm of ice and (200 - x) gm of water.
2. Steam condensed: the mass increase from 300 gm to 330 gm means 330 - 200 - 100 = 30 gm of steam condensed.
3. Heat lost by steam = 30 × [2250 + 4.2 × 50] = 30 × 2460 = 73800 J, using 22.5 × 10⁵ J/kg = 2250 J/gm.
4. Heat gained by calorimeter = \( \frac{100}{1000} \) × 0.42 × 10³ × (50 - 0) = 2100 J.
5. Heat gained by ice = \( \frac{x}{1000} \) × [3.36 × 10⁵ + 4.2 × 10³ × 50] = x × 546 J.
6. Heat gained by water = \( \frac{200 - x}{1000} \) × 4.2 × 10³ × 50 = 42000 - 210x J.
7. Heat gained = heat lost: 2100 + 546x + 42000 - 210x = 73800, so 336x = 29700 and x = 88.39 gm.
8. Ice = 88.39 gm and water = 200 - 88.39 = 111.61 gm, so the ratio of ice to water is 88.39 : 111.61 = 1 : 1.263 ≈ 0.79 (ice : water).

Teacher's Note:
a) The 30 gm of steam is found from the extra mass in the calorimeter, not from the steam given.
b) Melting ice uses 3.36 × 10⁵ J/kg, so 1 gm of ice absorbs 336 J.
c) Keep the units consistent: 1 gm is 10^-3 kg, so divide the masses by 1000 in each term.

Ex.4 A piece of metal weighs 46 gm in air. When it is immersed in a liquid of specific gravity 1.24 at 37ºC, it weighs 30 gm. When the temperature of the liquid is raised to 42ºC, the metal piece weighs 30.5 gm. The specific gravity of the liquid at 42ºC is 1.20. Calculate the coefficient of linear expansion of the metal.

Answer:
1. Loss of weight at the first reading (taken at 27ºC in the key): 46 - 30 = 16 gm, so the volume of metal V₂₇ = \( \frac{16}{1.24} \) cc = 12.90 cc.
2. Loss of weight at 42ºC: 46 - 30.5 = 15.5 gm, so the volume of metal V₄₂ = \( \frac{15.5}{1.20} \) cc = 12.92 cc.
3. Volume expansion: V₄₂ = V₂₇(1 + γT), with T = 42 - 27 = 15ºC.
4. 1 + 15γ = \( \frac{15.5}{1.20} \) × \( \frac{1.24}{16} \) = 1.00104, so γ = \( \frac{1}{15} \) × \( \frac{1}{960} \) = 6.94 × 10^-5 per ºC.
5. α = \( \frac{\gamma}{3} \) = \( \frac{1}{3 \times 15 \times 960} \) = 2.31 × 10^-5 per ºC.

Teacher's Note:
a) The question prints 37ºC for the first reading, but the key uses 27ºC; with 27ºC the rise is 15ºC and the standard answer is obtained.
b) If 37ºC were used, the rise would be only 5ºC and the result would be about 6.9 × 10^-5 per ºC, which is not a realistic value for a metal.
c) Volume expansion is three times linear expansion for an isotropic solid.

Ex.5 A one litre flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. What is the volume of mercury in flask ? Given coefficient of linear expansion of glass = 9 × 10 ^-6 per degree centrigrade. Coefficient of volume expansion of Hg = 1.8 × 10^-4 per degree centrigrade.

Answer:
1. Let x be the volume of mercury in the flask, and V = 1 litre = 1000 cm³.
2. Coefficient of cubical expansion of glass: γ_g = 3α = 3 × 9 × 10^-6 = 27 × 10^-6 per ºC.
3. Coefficient of cubical expansion of mercury: γ_m = 1.8 × 10^-4 per ºC.
4. The air space stays the same, so the expansion of the flask equals the expansion of mercury: V γ_g = x γ_m.
5. x = 1000 × \( \frac{27 \times 10^{-6}}{18 \times 10^{-5}} \) = 150 cm³.
6. Hence, the volume of mercury in the flask is 150 cm³.

Teacher's Note:
a) Use the cubical expansion of glass, which is 3α, not the linear coefficient.
b) The mercury expands much more than glass, so only a small volume of mercury is needed to match it.

Ex.6 An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and raised further by 44 cm. What will be the length of air column above mercury in the tube ? Atmospheric pressure = 76 cm of mercury.

Answer:
1. Let A be the area of cross-section. Initial air column = 8A at atmospheric pressure P₁ = 76 cm of Hg.
2. Let x be the new length of the air column. The mercury column inside the tube is 52 - x cm above the outside level, so P₂ = 76 - (52 - x) = (24 + x) cm of Hg.
3. New volume V₂ = xA.
4. Boyle's law: P₁V₁ = P₂V₂, so 76 × 8A = (24 + x) xA, which gives x² + 24x - 608 = 0.
5. x = \( \frac{-24 + \sqrt{(24)^2 + 4 \times 608}}{2} \) = 15.4 cm, the other root being negative.
6. Hence, the length of the air column is 15.4 cm.

Teacher's Note:
a) The pressure of the trapped air equals atmospheric pressure minus the mercury column above the outside level.
b) The key writes P₂ as (24 - x), but the equation used is (24 + x); the correct value is 24 + x.
c) Discard the negative root because a length cannot be negative.

Ex.7 An air bubble starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4 mm at the surface. The depth of the lake is 250 cm and the temperature at the surface is 40ºC. What is the temperature at the bottom of the lake ? Given atmospheric pressure = 76 cm of Hg and g = 980 cm/sec².

Answer:
1. Radius at the bottom: r₁ = 0.18 cm; radius at the surface: r₂ = 0.2 cm.
2. Pressure at the bottom: P₁ = (76 × 13.6 + 250 × 1) × 980 dyne/cm² = (1033.6 + 250) × 980 dyne/cm².
3. Pressure at the surface: P₂ = 76 × 13.6 × 980 dyne/cm² = 1033.6 × 980 dyne/cm².
4. Surface temperature: T₂ = 273 + 40 = 313 K.
5. Using \( \frac{P_1 V_1}{T_1} \) = \( \frac{P_2 V_2}{T_2} \) with V ∝ r³: T₁ = \( \frac{P_1}{P_2} \) × \( \left( \frac{r_1}{r_2} \right)^3 \) × T₂.
6. T₁ = \( \frac{1283.6}{1033.6} \) × (0.9)³ × 313 = 283.35 K.
7. Temperature at the bottom = 283.35 - 273 = 10.35ºC.

Volumes of the bubble (used in the ratio above): at the bottom V₁ = \( \frac{4}{3}\pi r_1^3 \) = \( \frac{4}{3}\pi (0.18)^3 \) cm³; at the surface V₂ = \( \frac{4}{3}\pi r_2^3 \) = \( \frac{4}{3}\pi (0.2)^3 \) cm³. Then \( \frac{(76 \times 13.6 + 250) 980 \times \frac{4}{3}\pi (0.18)^3}{T_1} \) = \( \frac{76 \times 13.6 \times 980 \times \frac{4}{3}\pi (0.2)^3}{313} \), i.e. \( \frac{1283.6 \times (0.18)^3}{T_1} \) = \( \frac{1033.6 \times (0.2)^3}{313} \).

Teacher's Note:
a) Water adds 1 gm/cm³ times g times depth to the pressure, so 250 cm of water gives 250 × 980 dyne/cm².
b) The volume of a bubble depends on the cube of its radius, which is why the small change in diameter matters.
c) Both temperatures must be in kelvin before using the gas law.

Ex.8 An electric bulb of volume 250 cm³ was sealed off during manufacture at the pressure of 10^-3 mm of Hg at 27ºC. Find the number of air molecules in the bulb.

Answer:
1. Given: V₁ = 250 cm³, P₁ = 10^-3 mm of Hg, T₁ = 273 + 27 = 300 K.
2. For N molecules, P₁V₁ = NkT, so 10^-3 × 250 = N × k × 300 ... (i)
3. At N.T.P., one mole of gas occupies V₀ = 22400 cm³ at P₀ = 760 mm of Hg and T = 273 K, and it contains N₀ = 6 × 10²³ molecules, so 760 × 22400 = N₀ × k × 273 ... (ii)
4. Dividing (i) by (ii): \( \frac{10^{-3} \times 250}{760 \times 22400} \) = \( \frac{N \times 300}{6 \times 10^{23} \times 273} \).
5. N = \( \frac{10^{-3} \times 250 \times 6 \times 10^{23} \times 273}{760 \times 22400 \times 300} \) = 8.02 × 10¹⁵ molecules.

Teacher's Note:
a) The answer is a count of molecules, not moles, so the key's unit "mole" is wrong.
b) Even a very low pressure bulb holds billions of molecules because the volume is large on the molecular scale.
c) The ratio method avoids needing the value of k directly.

Ex.9: One mole of argon expands polytropically, the polytropic constant being 1.5, that is, according to the law pV^1.5 = constant. In the process, its temperature changes by ΔT = - 26 K. Find

(a) the amount of heat obtained by the gas, and

Answer:
1. Argon is monatomic, so C_v = \( \frac{3}{2} \)R and γ = \( \frac{5}{3} \).
2. Change in internal energy: ΔU = nC_vΔT = 1 × \( \frac{3}{2} \) × 8.3 × (-26) = -324 J.
3. Work done by the gas: W = \( \frac{R \Delta T}{1 - 1.5} \) = -2R ΔT = -2 × 8.3 × (-26) = +432 J.
4. Heat absorbed: ΔQ = ΔU + W = -324 + 432 = 108 J.
5. Hence, the gas obtains 108 J of heat.

Teacher's Note:
a) For a polytropic process pV^x = constant, the work is \( \frac{nR\Delta T}{1 - x} \) and this sign is easy to get wrong.
b) The temperature falls, so the internal energy falls, yet the gas still gains heat in this process.

(b) the work performed by the gas.

Answer:
1. For pV^1.5 = constant with pV = RT, V ∝ 1/T², so p = \( \frac{RT}{V} \) ∝ T³.
2. W = \( \int p \, dV \) = -2R ΔT, which gives W = -2 × 8.3 × (-26) = 432 J.
3. Hence, the work performed by the gas is 432 J.

Derivation as in the source: ΔU = \( \int_T^{T+\Delta T} C_v \, dT \) = \( \frac{R}{\gamma - 1} \Delta T \) = \( \frac{8.3}{\frac{5}{3} - 1} \)(-26) = -324 J, and ΔW = \( \int_T^{T+\Delta T} p \, dV \). Since pV^1.5 = constant and pV = RT, V^0.5 ∝ \( \frac{1}{T} \), so V ∝ \( \frac{1}{T^2} \) or V = \( \frac{a}{T^2} \). Then p = \( \frac{RT}{V} \) = \( \frac{RT}{a/T^2} \) = \( \frac{R}{a} T^3 \), and ΔW = \( \int_T^{T+\Delta T} \frac{R}{a} T^3 \left( -\frac{2a}{T^3} \right) dT \) = -\( \int_T^{T+\Delta T} 2R \, dT \), so ΔW = -2RΔT = -2 × 8.3 × (-26) = +432 J. ΔQ = ΔU + ΔW = -324 + 432 = 108 J.

Teacher's Note:
a) Work is positive here because the gas expands, even though the temperature falls.
b) Check that ΔQ = ΔU + W gives 108 J, which matches part (a).

Ex.10: A composite rod is made by joining a copper rod end to end with a second rod of different material but of the same cross-section. At 25ºC, the composite rod is 1 m in length of which the length of the copper rod is 30 cm. At 125ºC, the length of the composite rod increases by 1.91 mm. When the composite rod is not allowed to expand by holding it between two rigid walls, it is found that the lengths of the two constituents do not change with the rise of temperature. Find the Young's modulus and the coefficient of linear expansion of the second rod. Given Young's modulus of copper = 1.3 × 10¹¹ N/m², coefficient of linear expansion of copper = 1.7 × 10^-5 / ºC.

Answer:
1. Length of copper rod l₁ = 30 cm, length of second rod l₂ = 70 cm.
2. Expansion of the composite rod for a rise of Δt = 100ºC: (l₁α₁ + l₂α₂)Δt = 0.191 cm.
3. (30 × 1.7 × 10^-5 + 70α₂) × 100 = 0.191, which gives α₂ = 2 × 10^-5 per ºC.
4. Since neither rod changes length when held between rigid walls, the thermal stresses must be equal: Y₁α₁ = Y₂α₂.
5. Y₂ = \( \frac{Y_1 \alpha_1}{\alpha_2} \) = \( \frac{1.3 \times 10^{11} \times 1.7 \times 10^{-5}}{2 \times 10^{-5}} \) = 1.1 × 10¹¹ N/m².

Teacher's Note:
a) The expansion equation uses each rod's length times its own coefficient.
b) Equal cross-sections mean the stresses are equal when the forces are equal, so area cancels.
c) The second rod expands more per degree than copper, so its Young's modulus comes out smaller than copper's.

Ex.11:A glass capillary tube, sealed at both ends is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube (which are equal in length) contain air at 27ºC and at a pressure of 76 cm of Hg. The tube is kept in a horizontal position such that the air column at one end is maintained at 0ºC and the other end at 127ºC. Calculate the length of the air column which is at 0ºC and its pressure. Neglect the change in volumes of mercury and glass.

[Figure: A horizontal tube of total length 100 cm with a 10 cm mercury column in the middle, and air on each side. The two air columns are each 45 cm long at the start, and the left air column is at 0ºC while the right air column is at 127ºC.]

Answer:
1. Initially each air column is 45 cm long, at P = 76 cm of Hg and T = 300 K.
2. Let x be the final length of the air column at 0ºC (T₁ = 273 K). The other air column is (90 - x) cm long at T₂ = 400 K.
3. The mercury is in equilibrium, so the pressures on both sides are equal: P₁ = P₂.
4. Using \( \frac{PV}{T} \) = constant for each side with equal pressures: \( \frac{Ax}{273} \) = \( \frac{A(90 - x)}{400} \).
5. 400x = 273(90 - x), so x = 36.5 cm.
6. Pressure: P = \( \frac{76 \times 45A \times 273}{300 \times 36.5A} \) = 85.27 cm of Hg.

Teacher's Note:
a) The mercury moves toward the colder end, which makes the cold air column shorter.
b) Use absolute temperatures, so 0ºC becomes 273 K and 127ºC becomes 400 K.
c) The final pressure is the same on both sides, which is the condition used to find x.

Ex.12 One gram-mole of oxygen at 27ºC and one atmospheric pressure is enclosed in a vessel.

(a) Assuming the molecules to be moving with v_r.m.s., find the number of collisions per second which the molecules make with one square metre area of the vessel wall.

Answer:
1. Number density: n = \( \frac{P}{kT} \) = \( \frac{1.01 \times 10^{5}}{1.38 \times 10^{-23} \times 300} \) = 2.44 × 10²⁵ m^-3.
2. v_rms = \( \sqrt{\frac{3RT}{M}} \) = \( \sqrt{\frac{3 \times 8.31 \times 300}{32 \times 10^{-3}}} \) = 483.4 m/s.
3. Only one-sixth of the molecules move towards a given face, so the number of collisions per second per m² = \( \frac{1}{6} \) n v_rms = \( \frac{1}{6} \) × 2.44 × 10²⁵ × 483.4.
4. Number of collisions = 1.97 × 10²⁷ per second per m².

Teacher's Note:
a) The one-sixth factor counts the molecules moving in the direction of one face.
b) Molar mass must be in kg, so 32 gm becomes 32 × 10^-3 kg.

(b) The vessel is next thermally insulated and moved with a constant speed v₀. It is then suddenly stopped. The process results in a rise of the temperature of the gas by 1ºC. Calculate the speed v₀.

Answer:
1. Kinetic energy of the moving gas, per mole: \( \frac{1}{2} \) M v₀².
2. When the vessel stops suddenly, this energy becomes internal energy, so \( \frac{1}{2} \) M v₀² = C_v × 1 K.
3. For oxygen, γ = 1.41, and C_v = \( \frac{R}{\gamma - 1} \).
4. v₀ = \( \sqrt{\frac{2R}{M(\gamma - 1)}} \) = \( \sqrt{\frac{2 \times 8.31}{32 \times 10^{-3} \times 0.41}} \) = 35.6 m/s.

Teacher's Note:
a) The temperature rise of 1 K is small, so the heat needed is simply C_v multiplied by 1.
b) The key's working writes 831 for R; the value used is R = 8.31 J/mol K, which gives 35.6 m/s.

Ex.13 A gaseous mixture enclosed in a vessel of volume V consists of one gram mole of a gas A with γ (C_p/C_v) = 5/3 and another has B with γ = 7/5 at a certain temperature T. The gram molecular weights of the gases A and B are 4 and 32 respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation PV^19/13 = constant, in adiabatic processes.

(a) Find the number of gram moles of the gas B in the gaseous mixture.

Answer:
1. For the mixture, γ_mix = \( \frac{19}{13} \) = 1 + \( \frac{R}{C_v} \), so \( \frac{R}{C_v} \) = \( \frac{6}{13} \) and C_v = \( \frac{13R}{6} \).
2. For gas A: C_v,A = \( \frac{3}{2} \)R. For gas B: C_v,B = \( \frac{5}{2} \)R.
3. Energy balance in an adiabatic change with no work at constant volume: (n_A + n_B)C_v = n_AC_v,A + n_BC_v,B.
4. With n_A = 1 gram mole: (1 + n_B) \( \frac{13}{6} \) = \( \frac{3}{2} \) + n_B \( \frac{5}{2} \).
5. 13 + 13n_B = 9 + 15n_B, so 2n_B = 4 and n_B = 2.
6. Hence, the mixture contains 2 gram moles of gas B.

Teacher's Note:
a) The mixture's C_v comes from its γ, using γ = 1 + R/C_v.
b) Use gram moles throughout, so the mole counts add directly.
c) The total number of moles is 3, which is needed in part (d).

(b) Compute the speed of sound in the gaseous mixture at T = 300 K.

Answer:
1. Mean molecular weight of the mixture: M = \( \frac{n_A M_A + n_B M_B}{n_A + n_B} \) = \( \frac{1 \times 4 + 2 \times 32}{1 + 2} \) = \( \frac{68}{3} \) gm = \( \frac{68}{3} \) × 10^-3 kg.
2. Speed of sound: v = \( \sqrt{\frac{\gamma RT}{M}} \) = \( \sqrt{\frac{19}{13} \times \frac{8.3 \times 300 \times 3}{68 \times 10^{-3}}} \).
3. v = 400.7 m/s.

Teacher's Note:
a) The mean molecular weight is a mole-weighted average, not a simple average of 4 and 32.
b) Sound speed depends on γ, R, T and M, so the mixture's γ = 19/13 is used here.

(c) It T is raised by 1 K from 300 K, find the percentage change in the speed of sound in the gaseous mixture.

Answer:
1. v ∝ √T, so \( \frac{v'}{v} \) = \( \sqrt{\frac{T'}{T}} \) = \( \sqrt{\frac{301}{300}} \).
2. Percentage change = \( \left( \sqrt{\frac{301}{300}} - 1 \right) \) × 100 = 0.17%.

Teacher's Note:
a) A 1 K rise near 300 K changes the speed by only about one-sixth of a percent.
b) The change is proportional to the square root of the temperature ratio, so it is half the fractional temperature change.

(d) The mixture is compressed adiabatically to 1/5th of its initial volume V. Find the change in its adiabatic compressibility in terms of the given quantities.

Answer:
1. Adiabatic compressibility: β = \( \frac{1}{\gamma P} \). Here the mixture has 3 gram moles, so P = \( \frac{3RT}{V} \).
2. Initial compressibility: β_i = \( \frac{V}{3\gamma RT} \) = \( \frac{13V}{57RT} \).
3. After compression to V/5, the pressure becomes P' = P × 5^19/13, so β_f = β_i × 5^-19/13.
4. Change: Δβ = β_f - β_i = -\( \frac{13V}{57RT} \) \( \left[ 1 - \frac{1}{5^{19/13}} \right] \).
5. Numerically, Δβ = -0.206 \( \frac{V}{RT} \), so the compressibility decreases.

In the form given in the source: Δβ = \( \frac{1}{\gamma P} \left[ 1 - \frac{1}{(5)^{\gamma}} \right] \) = \( \frac{13}{3 \times 19} \frac{V}{RT} \left[ 1 - \frac{1}{(5)^{\gamma}} \right] \), for which the source quotes 2.487 × 10^-3.

Teacher's Note:
a) The compressibility is the reciprocal of γ times pressure, so a higher pressure gives a smaller compressibility.
b) No numerical value of V or T is given, so the answer is left in terms of V/RT.

Ex.14 An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2.

(a) How many degrees of freedom do the gas molecules have ?

Answer:
1. For an adiabatic change, TV^γ-1 = constant, so T V^γ-1 = \( \frac{T}{2} \) (5.66 V)^γ-1.
2. (5.66)^γ-1 = 2, so γ - 1 = \( \frac{\log 2}{\log 5.66} \) = \( \frac{0.3010}{0.7528} \) = 0.4.
3. γ = 1.4, which is the value for a diatomic gas.
4. Hence, the molecules have 5 degrees of freedom.

Teacher's Note:
a) Use the adiabatic relation with T and V together, since both change.
b) γ = 1.4 corresponds to 5 degrees of freedom: 3 translational and 2 rotational.

(b) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V.

Answer:
1. Work in an adiabatic change: W = \( \frac{PV - P'V'}{\gamma - 1} \), where P' and V' are the final values.
2. Final pressure from P'V' / T' = PV / T: P' = \( \frac{PV}{T} \) × \( \frac{T/2}{5.66 V} \) = \( \frac{P}{11.32} \).
3. W = \( \frac{1}{0.4} \) \( \left[ PV - \frac{P}{11.32} \times 5.66V \right] \) = \( \frac{1}{0.4} \) \( \left[ PV - \frac{PV}{2} \right] \).
4. W = \( \frac{PV}{0.8} \) = 1.25 PV.

Teacher's Note:
a) The final pressure is found from the ideal gas law, using the temperature and volume ratios.
b) Work is positive because the gas expands.

Ex.15 An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature fall to T/2.

(a) How many degrees of freedom do the gas molecules have ?

Answer:
1. TV^γ-1 = \( \frac{T}{2} \) (5.66 V)^γ-1, so (5.66)^γ-1 = 2.
2. γ - 1 = \( \frac{\log 2}{\log 5.66} \) = 0.4, so γ = 1.4.
3. Hence, the molecules have 5 degrees of freedom, as for a diatomic gas.

Teacher's Note:
a) This is the same calculation as Ex.14(a), because the data are identical.
b) A diatomic gas at room temperature has 5 active degrees of freedom.

(b) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V.

Answer:
1. W = \( \frac{1}{1 - \gamma} \) [P₁V₁ - PV] in the notation of the adiabatic change, with γ = 1.4.
2. P₁ = \( \frac{T_1 V}{T V_1} \) P = \( \frac{P}{11.32} \) (since T₁ = T/2 and V₁ = 5.66 V).
3. W = \( \frac{1}{1 - 1.4} \) \( \left[ \frac{P}{11.32} \times 5.66 V - PV \right] \) = \( \frac{PV}{0.4} \) \( \left[ 1 - \frac{5.66}{11.32} \right] \).
4. W = \( \frac{PV}{0.4} \) × \( \frac{1}{2} \) = 1.25 PV.

Teacher's Note:
a) The sign of 1/(1 - γ) matches the expansion, so the work comes out positive.
b) The same answer 1.25 PV is found in Ex.14(b) by a different arrangement of the same terms.

Ex.16 Two moles of helium gas with γ equal to \( \frac{5}{3} \) are initially at temperature 27ºC and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.

(a) Sketch the process on P-V diagram.

Answer:
1. A to B is isobaric expansion at constant pressure, drawn as a horizontal line.
2. B to C is adiabatic expansion back to the initial temperature, drawn as a falling curve that is steeper than an isotherm.
3. P is in units of 10⁵ N/m² and V is in litres; the sketch is not to scale.

Teacher's Note:
a) An adiabat falls more steeply than an isotherm on a P-V diagram.
b) The curve ends at a lower pressure than the start, because the gas expands further.

(b) Find the final value of volume and pressure of the gas.

Answer:
1. Initial pressure: P₁ = \( \frac{nRT_1}{V_1} \) = \( \frac{2 \times 8.31 \times 300}{20 \times 10^{-3}} \) = 2.5 × 10⁵ N/m².
2. Isobaric step: \( \frac{T_2}{T_1} \) = \( \frac{V_2}{V_1} \) = 2, so T₂ = 600 K at V₂ = 40 × 10^-3 m³.
3. Adiabatic step from T₂ = 600 K back to T₃ = 300 K: \( \frac{V_3}{V_2} \) = \( \left( \frac{T_2}{T_3} \right)^{\frac{1}{\gamma - 1}} \) = 2^3/2 = 2.83.
4. Final volume: V₃ = 2.83 × 40 × 10^-3 = 0.113 m³ (113 litres).
5. Final pressure: P₃ = nRT₃/V₃ = \( \frac{2 \times 8.31 \times 300}{0.113} \) = 4.4 × 10⁴ N/m².

Teacher's Note:
a) The final temperature is the starting temperature, so the adiabatic step must bring the gas back to 300 K.
b) Pressure is found from the ideal gas law at the final state, which is quicker than the adiabatic relation.

(c) Find the work done by the gas under isobaric and adiabatic processes and total work done.

Answer:
1. Isobaric work: W_AB = P(V₂ - V₁) = 2.5 × 10⁵ × 20 × 10^-3 = 5000 J.
2. Adiabatic work: W_BC = \( \frac{nR(T_2 - T_3)}{\gamma - 1} \) = \( \frac{2 \times 8.31 \times 300}{\frac{2}{3}} \) = 7479 J.
3. Total work = 5000 + 7479 = 12479 J, which is about 1.25 × 10⁴ J.

Teacher's Note:
a) The key's total of 11479 J is an addition slip; 5000 + 7479 = 12479 J.
b) In the adiabatic step the gas cools, so its work comes from its internal energy.

Ex.17 Two moles of helium gas undergo a cyclic process as shown in the figure.

Assuming the gas to be ideal, calculate the following quantities in this process.

(a) The net change in the heat energy

Answer:
1. A to B is isobaric at 2 atm, from 300 K to 400 K: W_AB = nR(T_B - T_A) = 2 × 8.3 × 100 = 1660 J.
2. B to C is isothermal at 400 K, from 2 atm to 1 atm: W_BC = nRT ln 2 = 2 × 8.3 × 400 × 0.6931 = 4602.9 J.
3. C to D is isobaric at 1 atm, from 400 K to 300 K: W_CD = nR(T_D - T_C) = -1660 J.
4. D to A is isothermal at 300 K, from 1 atm to 2 atm: W_DA = nRT ln(P_D/P_A) = 2 × 8.3 × 300 × ln(1/2) = -3452.2 J.
5. Net work = 1660 + 4602.9 - 1660 - 3452.2 = 1150.7 J.
6. Net heat absorbed in a cycle equals the net work, so the net heat energy is 1150.7 J.

Teacher's Note:
a) In a closed cycle, the net heat and the net work are equal, because the internal energy returns to its start.
b) The isothermal steps use ln 2 = 0.6931, and the sign of each step depends on whether the gas expands or is compressed.

(b) The net work done

Answer:
1. Net work done = W_AB + W_BC + W_CD + W_DA = 1660 + 4602.9 - 1660 - 3452.2 J.
2. Net work done = 1150.7 J.

Teacher's Note:
a) The two isobaric steps cancel in work only when their temperature changes are equal in size and opposite in sign, as they are here.
b) The work is positive, so the cycle delivers net work.

(c) The net change in internal energy

Answer:
1. The gas returns to its original state after a cycle.
2. Internal energy depends only on the state, so the net change is ΔU = 0.

Teacher's Note:
a) Internal energy of an ideal gas depends only on temperature, which is the same at the start and end.
b) This is the reason the net heat equals the net work in part (a).

Ex.18 An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat involved in these steps are Q₁ = 5960 J, Q₂ = -5585 J, Q₃ = -2980 J and Q₄ = 3645 J respectively. The corresponding works involved are W₁ = 2200 J, W₂ = -825 J, W₃ = -1100 J and W₄ respectively.

(a) Find the value of W₄.

Answer:
1. Net heat absorbed: Q = Q₁ + Q₂ + Q₃ + Q₄ = 5960 - 5585 - 2980 + 3645 = 1040 J.
2. Net work: W = W₁ + W₂ + W₃ + W₄ = 2200 - 825 - 1100 + W₄ = 275 + W₄.
3. For a cycle, ΔU = 0, so Q = W, which gives 1040 = 275 + W₄.
4. W₄ = 765 J.

Teacher's Note:
a) Add the heats with their signs, since the negative values are heat given out.
b) The first law for a cycle reduces to net heat equals net work.

(b) What is the efficiency of the cycle ?

Answer:
1. Work done by the cycle = 275 + 765 = 1040 J.
2. Heat absorbed = Q₁ + Q₄ = 5960 + 3645 = 9605 J.
3. Efficiency η = \( \frac{1040}{9605} \) = 0.1083.
4. Hence, the efficiency is 10.83%.

Teacher's Note:
a) Only the positive heat inputs count as heat absorbed, so Q₂ and Q₃ are left out.
b) The efficiency 1040/9605 rounds to 10.83%.

Ex.19 At 27ºC, two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate

(a) the final temperature of the gas

Answer:
1. For an adiabatic change, T₁V^γ-1 = T₂(2V)^γ-1, with γ = \( \frac{5}{3} \) and T₁ = 300 K.
2. T₂ = 300 × \( \frac{1}{2^{2/3}} \) = 188.99 K.

Teacher's Note:
a) The gas cools when it expands adiabatically.
b) The exponent is γ - 1 = 2/3 for a monoatomic gas.

(b) change in its internal energy

Answer:
1. ΔU = nC_v(T₂ - T₁) = 2 × \( \frac{3}{2} \) × 8.3 × (188.99 - 300).
2. ΔU = 2 × 12.45 × (-111.01) = -2764 J.
3. The negative sign shows a fall in internal energy.

Teacher's Note:
a) The key's value of -2767.5 J is a small arithmetic slip; the correct value is -2764 J.
b) The change in internal energy is the same for any path between these two temperatures.

(c) the workdone by the gas during this process.

Answer:
1. For an adiabatic process, Q = 0, so the work done by the gas = -ΔU.
2. Work done by the gas = 2764 J.

Teacher's Note:
a) The gas does work at the expense of its internal energy, so the work done by the gas is positive.
b) The work done by the gas equals the fall in internal energy, so its magnitude is 2764 J.

Ex.20 A sample of 2 kg of monoatomic Helium (assumed ideal) is taken through the process ABC and another sample of 2 kg of the same gas is taken through the process ADC as shown in the figure. Given molecular mass of Helium = 4

(a) What is the temperature of Helium in each of the states A, B, C and D ?

Answer:
1. Number of moles: n = \( \frac{2 \times 10^{3}}{4} \) = 500.
2. State A: P_A = 5 × 10⁴ N/m², V_A = 10 m³, so T_A = \( \frac{P_A V_A}{nR} \) = \( \frac{5 \times 10^{4} \times 10}{500 \times 8.31} \) = 120.3 K.
3. State B: P_B = 10 × 10⁴ N/m², V_B = 10 m³, so T_B = 240.7 K.
4. State C: P_C = 10 × 10⁴ N/m², V_C = 20 m³, so T_C = 481.3 K.
5. State D: P_D = 5 × 10⁴ N/m², V_D = 20 m³, so T_D = 240.7 K.

Teacher's Note:
a) Each temperature follows from pV = nRT with the values read from the graph.
b) T_B and T_D are equal, which is a check that the two paths share the same pressure-volume product.

(b) Is there any way of telling afterwards which sample of Helium went through the process ABC and which went through the process ADC ? Write Yes or No.

Answer: No.

Teacher's Note:
a) Internal energy depends only on the state, and both samples start at A and end at C.
b) The heat and work differ between the paths, but these cannot be read from the final state.

(c) How much is the heat involved in each of the processes ABC and ADC ?

Answer:
1. Change in internal energy from A to C: ΔU = nC_vΔT = 500 × \( \frac{3}{2} \) × 8.3 × (481.3 - 120.3) = 2.25 × 10⁶ J, the same for both paths.
2. Path ABC: A to B is at constant volume (no work), and B to C is at constant pressure 10 × 10⁴ N/m², so W_ABC = 10⁵ × 10 = 1.0 × 10⁶ J.
3. Heat for ABC: Q_ABC = ΔU + W = 2.25 × 10⁶ + 1.0 × 10⁶ = 3.25 × 10⁶ J.
4. Path ADC: A to D is at constant pressure 5 × 10⁴ N/m², and D to C is at constant volume, so W_ADC = 5 × 10⁴ × 10 = 5.0 × 10⁵ J.
5. Heat for ADC: Q_ADC = 2.25 × 10⁶ + 0.5 × 10⁶ = 2.75 × 10⁶ J.

Teacher's Note:
a) Work is the area under each path on the P-V graph, and the isochoric legs do no work.
b) The heat is larger on ABC because the path does more work.

Ex.21 Two moles of an ideal monoatomic gas, initially at pressure p₁ and volume V₁ undergo an adiabatic compression until its volume is V₂. Then the gas is given heat Q at constant volume V₂.

(a) Sketch the complete process on a p-V diagram.

Answer:
1. A to B is the adiabatic compression from V₁ to V₂, drawn as a steep curve rising to the left.
2. B to C is heating at constant volume V₂, drawn as a vertical line upward.
3. The pressure at C is p₃, which is higher than p₂.

Teacher's Note:
a) The key labels BC as isobaric heating, but heating at constant volume is isochoric, so BC is a vertical line.
b) The adiabat is steeper than an isotherm, which is why p₂ is higher than p₁ at the smaller volume.

(b) Find the total workdone by the gas, the total change in its internal energy and the final temperature of the gas.

[Give your answers in terms of p₁, V₁, V₂, Q and R].

Answer:
1. Initial temperature: T₁ = \( \frac{p_1 V_1}{2R} \). For adiabatic compression, T₂ = T₁ \( \left( \frac{V_1}{V_2} \right)^{2/3} \) = \( \frac{p_1 V_1^{5/3}}{2R V_2^{2/3}} \).
2. Work by the gas in the adiabatic step: W_AB = \( \frac{nR(T_1 - T_2)}{\gamma - 1} \) = \( \frac{3}{2} p_1 V_1 \left[ 1 - \left( \frac{V_1}{V_2} \right)^{2/3} \right] \).
3. No work is done in the constant volume step, so the total work by the gas is W = \( \frac{3}{2} p_1 V_1 \left[ 1 - \left( \frac{V_1}{V_2} \right)^{2/3} \right] \), which is negative because the gas is compressed.
4. Total change in internal energy: ΔU = Q - W = Q + \( \frac{3}{2} p_1 V_1 \left[ \left( \frac{V_1}{V_2} \right)^{2/3} - 1 \right] \).
5. Final temperature: T₃ = T₂ + \( \frac{Q}{nC_v} \) = T₂ + \( \frac{Q}{3R} \), so T₃ = \( \frac{p_1 V_1}{2R} \left( \frac{V_1}{V_2} \right)^{2/3} + \frac{Q}{3R} \).

Teacher's Note:
a) The adiabatic step gives the work and the change in internal energy, and the heating step only changes internal energy.
b) The heating of BC is at constant volume V₂, so it is isochoric and not isobaric.
c) Use n = 2 moles so that nC_v = 3R in the final temperature step.

Ex.22 A bar of copper of length 75 cm and a bar of steel of length 125 cm are joined together end to end. Both are of circular cross-section with diameter 2 cm. The free ends of copper and steel are maintained at 100ºC and 0ºC, respectively. The surfaces of the bars are thermally insulated. What is the temperature of the copper-steel junction ? What is the heat transmitted per unit time across the junction ? Thermal conductivity of copper is 9.2 × 10^-2 kilocal/mºC.s and that of steel is 1.1 × 10^-2 kilocal/mºC.s.

What is the temperature of the copper-steel junction ?

Answer:
1. In the steady state, the heat flowing per second through both bars is the same: \( \frac{K_1 A (100 - T)}{l_1} \) = \( \frac{K_2 A (T - 0)}{l_2} \).
2. Solving for the junction temperature: T = \( \frac{\frac{K_1}{l_1} \times 100}{\frac{K_1}{l_1} + \frac{K_2}{l_2}} \) = \( \frac{\frac{9.2}{0.75} \times 100}{\frac{9.2}{0.75} + \frac{1.1}{1.25}} \).
3. T = \( \frac{1226.7}{12.27 + 0.88} \) = \( \frac{1226.7}{13.15} \) = 93.3ºC.

Teacher's Note:
a) Lengths must be in metres to match the conductivity units per metre.
b) The junction is much closer to 100ºC, because copper is the better conductor.

What is the heat transmitted per unit time across the junction ?

Answer:
1. Area of cross-section: A = π(0.01)² m² = 3.14 × 10^-4 m².
2. Heat per second through copper: Q = \( \frac{K_1 A (100 - T)}{l_1} \) = \( \frac{9.2 \times 10^{-2} \times 3.14 \times 10^{-4} \times (100 - 93.3)}{0.75} \).
3. Q = 2.6 × 10^-4 kilocal/s.

Teacher's Note:
a) The key's working writes (100 - 933); the junction temperature used is 93.3ºC.
b) The answer is a very small heat flow, since the bars are thin and the conductivities are small.

Ex.23 A cylindrical vessel of 28 cm diameter contains 20 g of nitrogen compressed by a piston supporting a weight of 75 kg. The temperature of the gas is 17° C. What work will the gas do, if it is heated to a temperature of 250° C? What amount of heat should be supplied? To what distance will the weight be raised? The process should be assumed to be isobaric; the heating of the vessel and the external pressure are negligible.

What work will the gas do, if it is heated to a temperature of 250° C?

Answer:
1. Work done in an isobaric change: W = \( \frac{m}{M} \) R ΔT = \( \frac{20}{28} \) × 8.3 × (250 - 17) = 1381 J.
2. Hence, the gas does 1381 J of work.

Teacher's Note:
a) Use the temperature difference in kelvin, which is the same as in degrees Celsius.
b) Nitrogen has 28 gm per mole, so 20 g is \( \frac{20}{28} \) of a mole.

What amount of heat should be supplied?

Answer:
1. Change in internal energy: ΔU = \( \frac{m}{M} \) C_v ΔT = \( \frac{20}{28} \) × \( \frac{R}{\gamma - 1} \) × 233, with γ = \( \frac{7}{5} \).
2. ΔU = \( \frac{20}{28} \) × \( \frac{8.3}{0.4} \) × 233 = 3453 J.
3. Heat supplied: ΔQ = ΔW + ΔU = 1381 + 3453 = 4834 J.

Teacher's Note:
a) Nitrogen is diatomic, so γ = 1.4 is used for C_v.
b) Heat supplied is larger than the work because part of it raises the internal energy.

To what distance will the weight be raised?

Answer:
1. The work done by the gas lifts the weight, so W = mgh.
2. 1381 = 75 × 9.8 × h, so h = \( \frac{1381}{735} \) = 1.88 m.

Teacher's Note:
a) The weight is lifted by the work done, not by the heat supplied.
b) Use g = 9.8 m/s² for the weight of 75 kg.

Ex.24 The temperature of air above a lake is -10ºC. At some instant, the thickness of ice in the lake is 2 cm. Calculate the time required for the thickness to be doubled. Thermal conductivity of ice = 0.004 cal/cm/s/ºC, density of ice = 0.92 g/cm³ and latent heat of ice = 80 cal/g.

Answer:
1. Let the ice thickness be x at some instant, with A the area and θ = 10ºC the air temperature below 0ºC.
2. Heat to freeze a layer of thickness dx: dQ = A ρ dx L.
3. Heat conducted through the ice in time dt: dQ = \( \frac{KA\theta}{x} \) dt.
4. Equating: A ρ L dx = \( \frac{KA\theta}{x} \) dt, so dt = \( \frac{\rho L}{K\theta} \) x dx.
5. Integrating from x₁ = 2 cm to x₂ = 4 cm: t = \( \frac{\rho L}{2K\theta} \) (x₂² - x₁²) = \( \frac{0.92 \times 80}{2 \times 0.004 \times 10} \) (16 - 4).
6. t = 11040 s, which is about 3.07 hours.

Teacher's Note:
a) The growing ice layer means the conduction path gets longer, so the rate slows down.
b) The time depends on the square of the thickness, which is why doubling the thickness takes much longer than the first 2 cm.

Ex.25 A rubber tube of length 20 cm, through which steam at 100ºC is passing is immersed in a calorimeter whose water equivalent is 15 gm and which contains 300 gm of water at 16ºC. The temperature of water rises at the rate of 2ºC per minute. If the outer and inner diameters of the tube are 1.0 cm and 0.6 cm respectively, calculate the thermal conductivity of rubber.

Answer:
1. Rate of heat gain of the water and calorimeter: Q = (300 + 15) × \( \frac{2}{60} \) = 10.5 cal/s.
2. Radial heat flow through the tube: Q = \( \frac{K \times 2\pi l (T_1 - T_2)}{\log_e (r_2 / r_1)} \), with r₂ = 0.5 cm, r₁ = 0.3 cm, l = 20 cm, T₁ = 100ºC and T₂ = 16ºC.
3. log_e(r₂/r₁) = log_e(1.667) = 0.511.
4. K = \( \frac{Q \log_e (r_2 / r_1)}{2\pi l (T_1 - T_2)} \) = \( \frac{10.5 \times 0.511}{2 \times 3.14 \times 20 \times 84} \) = 5.1 × 10^-4 cal cm^-1 s^-1 ºC^-1.

In general form (as in the source): Q = (m + w) dT cal/sec ... (ii). Equating (i) and (ii), \( \frac{K \cdot 2\pi l (T_1 - T_2)}{\log_e (r_2 / r_1)} \) = (m + w) dT, so K = \( \frac{(m + w) dT \cdot \log_e (r_2 / r_1)}{2\pi l (T_1 - T_2)} \). Substituting, K = \( \frac{(300 + 15)(2/60) \times 2.3 \log_{10}(1.0/0.6)}{2 \times 3.14 \times 20 \times (100 - 16)} \) = \( \frac{(315)(1/30) \times (2.3 \times 0.2219)}{2 \times 3.14 \times 20 \times 84} \) = 5 × 10^-4 cal cm^-1 s^-1 ºC^-1.

Teacher's Note:
a) Radii are half the diameters, and only the ratio r₂/r₁ enters the logarithm.
b) The rate of rise is per minute, so divide by 60 to get degrees per second.

Ex.26 A cylindrical block of length 0.4 m and area of cross-section 0.04 m² is placed coaxially on a thin metallic disc of mass 0.4 kg and of the same cross-section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the material of the cylinder is 100 watt/m-K and the specific heat of the material of the disc is 600 J/kg-K, how long will it take for the temperature of the disc to increase to 350 K ? Assume, for the purpose of calculation, the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder.

Answer:
1. Rate of heat flow through the cylinder at disc temperature T: \( \frac{dQ}{dt} \) = \( \frac{KA(400 - T)}{L} \).
2. Heat used to raise the disc: \( \frac{dQ}{dt} \) = ms \( \frac{dT}{dt} \).
3. Equating: dt = \( \frac{msL}{KA} \) \( \frac{dT}{400 - T} \).
4. Integrating from T = 300 K to 350 K: t = \( \frac{msL}{KA} \) ln \( \frac{400 - 300}{400 - 350} \) = \( \frac{0.4 \times 600 \times 0.4}{100 \times 0.04} \) × ln 2.
5. t = 24 × 0.6931 = 16.6 s.

Teacher's Note:
a) The key uses 10 W/m K, but the question gives 100 W/m K; with 100 the time is 16.6 s, not 166 s.
b) The disc temperature rises faster at first, when the gap to 400 K is large.

Ex.27 A liquid placed in a container open to atmosphere takes 5 minutes to cool from 80ºC to 50ºC. How much time will it take to cool from 60ºC to 30ºC ? The temperature of the surroundings is 20ºC.

Answer:
1. Newton's law of cooling: \( \frac{dT}{dt} \) = -K(T - T₀), so \( \log_e \frac{T_2 - T_0}{T_1 - T_0} \) = -Kt.
2. First case: 5 min, from 80ºC to 50ºC: \( \log_e \frac{30}{60} \) = -5K, so 5K = log_e 2.
3. Second case: from 60ºC to 30ºC in time t: \( \log_e \frac{10}{40} \) = -Kt, so Kt = log_e 4.
4. Dividing: \( \frac{t}{5} \) = \( \frac{\log_e 4}{\log_e 2} \) = 2, so t = 10 minutes.

Teacher's Note:
a) The surroundings at 20ºC set the reference temperature for the cooling law.
b) The cooling constant K is the same in both cases because the container and liquid do not change.

Ex.28 A solid copper sphere cools at the rate of 2.8ºC per minute, when its temperature is 127ºC. Find the rate at which another copper sphere of twice the radius will lose its temperature at 327ºC, if in both the cases, the room temperature is maintained at 27ºC.

Answer:
1. Rate of loss of heat by radiation: \( \frac{dQ}{dt} \) = msdT/dt = σA(T⁴ - T₀⁴), so \( \frac{dT}{dt} \) = \( \frac{\sigma A}{ms} \)(T⁴ - T₀⁴).
2. For a sphere of radius r: m = \( \frac{4}{3} \) πr³ρ and A = 4πr², so \( \frac{dT}{dt} \) = \( \frac{3\sigma}{r \rho s} \)(T⁴ - T₀⁴).
3. For the first sphere at 400 K and room temperature 300 K: 2.8 = \( \frac{3\sigma}{r\rho s} \)(400⁴ - 300⁴) ... (i)
4. For the sphere of radius 2r at 600 K: \( \frac{dT}{dt} \) = \( \frac{3\sigma}{2r\rho s} \)(600⁴ - 300⁴) ... (ii)
5. Dividing (ii) by (i): \( \frac{dT}{dt} \) = 2.8 × \( \frac{1}{2} \) × \( \frac{6^4 - 3^4}{4^4 - 3^4} \) = 2.8 × \( \frac{1}{2} \) × \( \frac{1215}{175} \) = 9.72 ºC per minute.

Teacher's Note:
a) Doubling the radius doubles the heat capacity but also doubles the area, so the ratio depends on the radius only through the 1/r factor.
b) Use absolute temperatures in the fourth power terms, here 400 K, 600 K and 300 K.

Ex.29 Energy falling on 1.0 m² area placed at right angles to a sun beam just outside the earth's atmosphere is 1.35 K joule in one second. Find sun's surface temperature. Mean distance of earth from sunis 1.50 × 10⁸ km, mean distance of sun = 1.39 × 10⁶ km and Stefan's constant = 5.67 × 10^-8 watt m^-2 K^-4.

Answer:
1. Solar constant: S = 1.35 kJ/m²s = 1.35 × 10³ W/m².
2. Power radiated per unit area of the sun: σT⁴, and at the earth's distance r this is spread over a sphere of radius r, so S = σT⁴ (R/r)², where R is the sun's radius.
3. T⁴ = \( \frac{S}{\sigma} \) (r/R)² = \( \frac{1.35 \times 10^{3}}{5.67 \times 10^{-8}} \) × (215.8)² = 2.38 × 10¹⁰ × 46570 = 1.108 × 10¹⁵ K⁴.
4. Here r/R = \( \frac{1.50 \times 10^{8}}{0.695 \times 10^{6}} \) = 215.8.
5. T = 5.77 × 10³ K = 5770 K.

Teacher's Note:
a) The quoted "mean distance of sun" of 1.39 × 10⁶ km is the diameter of the sun, so the radius 0.695 × 10⁶ km is used.
b) The sun's surface temperature is found from the radiation law, not from the distance alone.

Ex.30 Two bodies A and B have thermal emissivity of 0.1 and 0.81 respectively. The outer surface areas of the two bodies are identical. These two bodies emit total radiative power at the same rate. The wavelength λ_B corresponding to the maximum spectral radiancy in the radiation from B is 1.0 μm larger than the wavelength λ_A corresponding to the maximum spectral radiancy in the radiation from A. If the temperature of body A is 5802 K, find

(a) temperature of (B) and

Answer:
1. Power radiated: P = eσAT⁴. Since the powers are equal and the areas are the same, e_AT_A⁴ = e_BT_B⁴.
2. 0.1 × T_A⁴ = 0.81 × T_B⁴, so \( \left( \frac{T_B}{T_A} \right)^4 \) = \( \frac{0.1}{0.81} \) = \( \frac{1}{81} \).
3. T_B = \( \frac{T_A}{3} \) = \( \frac{5802}{3} \) = 1934 K.

Teacher's Note:
a) Equal radiated power with equal areas leaves a ratio of emissivity times the fourth power of temperature.
b) Since 81 is 3⁴, the fourth root is exactly 3.

(b) λ_B.

Answer:
1. Wien's law: λ_mT = constant, so λ_AT_A = λ_BT_B, giving \( \frac{\lambda_B}{\lambda_A} \) = \( \frac{T_A}{T_B} \) = 3.
2. λ_A = \( \frac{\lambda_B}{3} \).
3. λ_B - λ_A = 1.0 μm, so λ_B - \( \frac{\lambda_B}{3} \) = \( \frac{2\lambda_B}{3} \) = 1.0 μm.
4. Hence, λ_B = 1.5 μm.

Teacher's Note:
a) Wien's displacement law gives the wavelength at maximum emission, which is inversely proportional to temperature.
b) The wavelength of body A is 0.5 μm, which lies in the visible range near blue-green light.

Heat gained by 250 gm of water and 50 gm of water equivalent of calorimeter at 100ºC to 0ºC
= 250 × 1 × (100 – 0) + 50 × (100 – 0)
= 300 × 100 = 30000 cal.
Total heat gained = 30000 cal. + 36000 = 66000 cal. .....(ii)
Amount of heat lost by steam (i) is greater than heat gained by ice. This shows that only a part of the steam will condense to water at 100ºC which will be sufficient for melting ice.
Let M be mass of steam which will be sufficient to melt ice,
∴ Mass M of steam required is given by M × 540 = 66000
or M = 66000/540 = 9
1100 = 122.2 gm.
Final temperature of system = 100ºC
Weight of contents = weight of ice + water + steam condensed
= 250 + 200 + 122.2
= 572.2 gm.