CBSE Class 12 Physics Heat and Thermodynamics Basic Problems

CBSE Class 12 Physics Heat and Thermodynamics Basic Problems. 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.

CLASS 12 PHYSICS : HEAT AND THERMODYNAMICS
BASIC PROBLEMS

1. A cylinder closed at both ends is separated into two equal (42 cm each) parts by a piston impermeable to heat. Both the parts contain the same masses of a gas at a temperature of 27° C and a pressure of 1 atm. How much should the gas in one part of the cylinder be heated to shift the piston by 2 cm? Find the pressure of the gas after the shifting of the piston.

Answer:
1. Initial temperature T = 27 + 273 = 300 K, pressure p = 1 atm, each part has length 42 cm.
2. After the shift, the heated part has length 44 cm and the other part has length 40 cm.
3. The piston does not conduct heat, so the other gas stays at 300 K. By Boyle's law, 1 × 42 = p' × 40, so p' = 1.05 atm.
4. The pressures on both sides of the piston are equal, so the heated gas is also at 1.05 atm.
5. For the heated gas, p'V'/T' = pV/T, so T' = 300 × (1.05 × 44)/(1 × 42) = 330 K = 57°C.
6. Hence, the gas must be heated to 57°C (a rise of 30°C), and the final pressure is 1.05 atm.

Teacher's Note:
a) The unheated gas is compressed isothermally, because the piston lets no heat through.
b) Equal pressure on both sides of the piston is the key condition for equilibrium.
c) Convert to kelvin before using the gas law.

2. A tall cylindrical vessel with gaseous nitrogen is located in the gravitational field. The temperature of the gas varies with height h in such a way that its density is the same throughout the volume. Find the temperature gradient.

Answer:
1. For hydrostatic equilibrium, dp/dh = -ρg.
2. For an ideal gas p = ρRT/M, and ρ is constant, so dp/dh = (ρR/M)(dT/dh).
3. Equating, (ρR/M)(dT/dh) = -ρg.
4. So dT/dh = -gM/R.
5. For nitrogen M = 28 g/mol, so dT/dh = -(9.8 × 0.028)/8.31 = -0.033 K/m, a fall of about 3.3 K per 100 m.

Teacher's Note:
a) Start from the pressure change in a fluid column and the ideal gas equation together.
b) The negative sign shows that the temperature falls with height.

3. A horizontal cylinder closed at one end is rotated with a constant angular velocity ω about a vertical axis passing through the open end of the cylinder. The outside pressure is p₀, the absolute temperature is T and the molar mass of air is M. Find the air pressure as a function of the distance r from the axis of rotation. The molar mass is assumed to be independent of r.

Answer:
1. Consider a thin layer of air of thickness dr at distance r. The pressure difference across it provides the centripetal force: dp = ρω²r dr.
2. For an ideal gas at constant temperature, ρ = pM/RT.
3. So dp/p = (Mω²/RT) r dr.
4. Integrating from r = 0 (where p = p₀) to r: ln(p/p₀) = Mω²r²/(2RT).
5. Hence, p = p₀ e^Mω²r²/2RT.

Teacher's Note:
a) The air at the open end is in contact with the outside, so p = p₀ at r = 0.
b) The pressure rises exponentially with r, because the density increases with pressure.
c) The result is similar to the barometric formula with gh replaced by ω²r²/2.

4. An ideal gas at 300 K occupies a volume of 0.5 m³ at a pressure of 2 atm. The gas expands adiabatically until its volume is 1.2 m³. Next, the gas is compressed isobarically to its original volume. Finally, the pressure is increased isochorically until the gas returns to its initial state. Represent the change on the p-V diagram. Determine the temperature at the end of each transformation, and the work done during the cycle. Assume γ = 1.4.

Answer:
1. Let the states be A (2 atm, 0.5 m³, 300 K), B (after adiabatic expansion, 1.2 m³) and C (after isobaric compression, 0.5 m³).
2. On the p-V diagram, A to B is a falling curve, B to C is a horizontal line to the left at the lower pressure, and C to A is a vertical line upwards.
3. T_B: TV^γ-1 = constant, so T_B = 300 × (0.5/1.2)^0.4 = 211.3 K.
4. Pressure at B: p_B = 2 × (0.5/1.2)^1.4 = 0.587 atm.
5. T_C: at constant pressure T is proportional to V, so T_C = 211.3 × (0.5/1.2) = 88.1 K.
6. W_AB = (p_AV_A - p_BV_B)/(γ - 1) = (1.013 × 10⁵ - 0.714 × 10⁵)/0.4 = 7.5 × 10⁴ J.
7. W_BC = p_B(V_C - V_B) = 0.587 × 1.013 × 10⁵ × (0.5 - 1.2) = -4.2 × 10⁴ J.
8. W_CA = 0 because the volume is constant.
9. Net work done in the cycle = 7.5 × 10⁴ - 4.2 × 10⁴ + 0 = 3.3 × 10⁴ J.

Teacher's Note:
a) Convert 1 atm to 1.013 × 10⁵ Pa before finding the work.
b) Work in an isochoric step is zero.
c) Net work in the cycle equals the area enclosed on the p-V diagram.

5. Five moles of neon gas (molecular weight = 20) at 2 atm and 27° C is adiabatically compressed to one-third its initial volume. Find the final pressure, the temperature and the work done on the gas.

Answer:
1. Neon is monoatomic, so γ = 5/3 and C_v = (3/2)R. Initial T₁ = 300 K.
2. p₂ = p₁(V₁/V₂)^γ = 2 × 3^5/3 = 12.5 atm.
3. T₂ = T₁(V₁/V₂)^γ-1 = 300 × 3^2/3 = 624 K = 351°C.
4. Work done on the gas = nC_v(T₂ - T₁) = 5 × 1.5 × 8.31 × 324 = 2.02 × 10⁴ J = 20.2 kJ.
5. Hence, the final pressure is 12.5 atm, the temperature is about 351°C and the work done on the gas is 20.2 kJ.

Teacher's Note:
a) In an adiabatic process all the work done on the gas goes into its internal energy.
b) Convert the final temperature back to °C only at the end.

6. One mole of certain ideal gas is contained under the weightless piston of a vertical cylinder at a temperature T. The space over the piston opens into the atmosphere of pressure p₀. What work has to be done in lifting the piston so that the volume of the gas under the piston increases isothermally n times?

Answer:
1. The piston is weightless, so initially the gas pressure equals p₀ and p₀V₁ = RT.
2. Work done by the gas in the isothermal expansion = RT ln n.
3. Work done by the gas against the atmosphere = p₀(nV₁ - V₁) = (n - 1)RT.
4. The gas pressure falls below p₀, so the outside agent must supply the difference: work A = (n - 1)RT - RT ln n.

Teacher's Note:
a) The agent's work equals work against the atmosphere minus work done by the gas.
b) For n greater than 1, (n - 1) is greater than ln n, so the answer is positive.

7. As a result of the isobaric heating by ΔT = 72 K one mole of a certain ideal gas receives heat Q = 1.6 kJ. Find the work performed by the gas, the increment of its internal energy and the value of its adiabatic exponent γ.

Answer:
1. For an isobaric process the work done by one mole is W = RΔT = 8.3 × 72 = 597.6 J.
2. By the first law, ΔU = Q - W = 1600 - 597.6 = 1002.4 J.
3. ΔU = C_vΔT, so C_v = 1002.4/72 and Q = C_pΔT.
4. γ = C_p/C_v = Q/ΔU = 1600/1002.4 = 1.6.

Teacher's Note:
a) For one mole at constant pressure, W = RΔT.
b) The ratio of heat to the rise in internal energy gives γ directly.

8. A reversible heat engine carries 1 mole of an ideal monoatomic gas around the cycle 1-2-3-1. Process 1-2 takes place at constant volume, process 2-3 is adiabatic and process 3-1 takes place at constant pressure. Compute the values for the heat Δ Q, the change in internal energy ΔU, and the work done ΔW, for each of the three processes and for the cycle as whole.

Answer:
1. The monoatomic gas has C_v = (3/2)R and C_p = (5/2)R; take R = 8.3 J/mol K.
2. Process 1-2 (constant volume, 300 K to 600 K): ΔW₁₂ = 0, ΔU₁₂ = (3/2) × 8.3 × 300 = 3735 J, ΔQ₁₂ = ΔU₁₂ = 3735 J.
3. Process 2-3 (adiabatic, 600 K to 455 K): ΔQ₂₃ = 0, ΔU₂₃ = (3/2) × 8.3 × (455 - 600) = -1805 J, ΔW₂₃ = -ΔU₂₃ = 1805 J.
4. Process 3-1 (constant pressure, 455 K to 300 K): ΔU₃₁ = (3/2) × 8.3 × (300 - 455) = -1930 J, ΔQ₃₁ = (5/2) × 8.3 × (-155) = -3216 J, ΔW₃₁ = ΔQ - ΔU = -3216 + 1930 = -1286 J.
5. For the whole cycle: ΔU = 3735 - 1805 - 1930 = 0, ΔQ = 3735 + 0 - 3216 = 519 J and ΔW = 0 + 1805 - 1286 = 519 J.
6. Hence, ΔQ = ΔW = 519 J and ΔU = 0 for the cycle.

Teacher's Note:
a) For a complete cycle ΔU = 0, so net heat absorbed equals net work done.
b) Check each step with the first law, ΔQ = ΔU + ΔW.
c) A negative ΔQ means that heat is rejected by the gas.

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
(b) the work performed by the gas.

Answer:
1. Argon is monoatomic, so C_v = (3/2)R. For a polytropic process pVⁿ = constant with n = 1.5, the work is W = RΔT/(1 - n).
2. (b) W = 8.31 × (-26)/(1 - 1.5) = 432 J.
3. Change in internal energy ΔU = C_vΔT = 1.5 × 8.31 × (-26) = -324 J.
4. (a) Q = ΔU + W = -324 + 432 = 108 J.
5. Hence, heat obtained = 108 J and work performed by the gas = 432 J.

Teacher's Note:
a) The work is positive even though the temperature falls, because the gas expands.
b) Use the first law after finding ΔU and W separately.

10. An ideal gas expands adiabatiacally from an initial temperature T₁ to a final temperature T₂. Prove that the work done by the gas is C_v(T₁ - T₂).

Answer:
1. Consider one mole of the gas. In an adiabatic process no heat is exchanged, so Q = 0.
2. By the first law of thermodynamics, Q = ΔU + W, so W = -ΔU.
3. For an ideal gas, the change in internal energy is ΔU = C_v(T₂ - T₁), whatever the process.
4. So W = -C_v(T₂ - T₁) = C_v(T₁ - T₂).
5. Hence proved. For T₂ less than T₁, the work done by the gas is positive.

Teacher's Note:
a) The internal energy of an ideal gas depends only on temperature.
b) For n moles multiply the result by n.

11. Ten grams of oxygen is heated at constant atmospheric pressure from 27°C to 127°C. How much heat is transferred to the oxygen? What fraction of the heat is used to increase the internal energy of the oxygen.

Answer:
1. Number of moles n = 10/32 = 0.3125 mol. Oxygen is diatomic, so C_p = (7/2)R and C_v = (5/2)R.
2. Heat transferred Q = nC_pΔT = 0.3125 × 3.5 × 8.31 × 100 = 909 J.
3. Increase in internal energy ΔU = nC_vΔT = 0.3125 × 2.5 × 8.31 × 100 = 649 J.
4. Fraction used for internal energy = ΔU/Q = C_v/C_p = 5/7 = 0.71.
5. Hence, the heat transferred is about 909 J and the fraction is 0.71.

Teacher's Note:
a) The rest of the heat, about 29 percent, is the work done by the gas in expanding.
b) Fraction ΔU/Q at constant pressure is 1/γ.

12. Three rods of material x and two rods of material y are connected as shown in the figure here. All the rods are identical (each 1 m long and of cross-sectional area 100 cm²). At the junctions A and C, there are two sources of heat of 25 W and 100 W, and at B and at D there are two sinks maintained at 30°C and 10°C, respectively. Calculate the temperatures of the points A and C is the steady stae. The thermal conductivity of x is 200 J s^-1 m^-1K^-1 and that of y is 400 J s^-1 m^-1 K^-1.

Answer:
1. Thermal conductance of a rod = KA/L. For x: 200 × 0.01/1 = 2 W/K. For y: 400 × 0.01/1 = 4 W/K.
2. The rods are AB, AD and AC (all x), and CB, CD (both y).
3. In the steady state, heat given at A flows out through AB, AD and AC: 25 = 2(T_A - 30) + 2(T_A - 10) + 2(T_A - T_C), which gives 6T_A - 2T_C = 105.
4. At C the heat given plus the heat from A flows out through CB and CD: 100 + 2(T_A - T_C) = 4(T_C - 30) + 4(T_C - 10), which gives 10T_C - 2T_A = 260.
5. Solving the two equations: T_A = 28°C and T_C = 31.6°C approximately (28.04°C and 31.6°C).

Teacher's Note:
a) Write one heat balance equation for each junction, treating conductances like electrical conductances.
b) Heat flows from the hotter C to the cooler A through the middle rod.
c) Check: 6 × 28.04 - 2 × 31.6 = 105 and 10 × 31.6 - 2 × 28.04 = 260.

13. A cylindrical block of length l = 0.4 m and area of cross-section A = 0.04 m² is placed coaxially on a thin metal disc of mass m = 0.4 kg of the same cross section. The upper face of the cylinder is maintained at a constant temperature T = 400 K and the initial temperature of the disc is T₀ = 300 K. If the thermal conductivity of the material of the cylinder is λ = 10 W m^-1K^-1 and the specific heat of the material of the disc is s = 600 J kg^-1 - K^-1, how long will it take for the temperature of the disc to increase to T₁ = 350 K? Assume for purposes of calculation that the thermal conductivity is very high and that the system is thermally insulated except for the upper face of the cylinder.

Answer:
1. Let the disc be at temperature T' at some instant. Heat flows through the cylinder at the rate λA(T - T')/l.
2. This heat raises the temperature of the disc: ms dT'/dt = λA(T - T')/l.
3. So dT'/(T - T') = (λA/(msl)) dt. Here λA/(msl) = (10 × 0.04)/(0.4 × 600 × 0.4) = 0.4/96 = 1/240 s⁻¹.
4. Integrating from T₀ to T₁: ln[(T - T₀)/(T - T₁)] = t/240.
5. t = 240 × ln(100/50) = 240 × 0.693 = 166.4 s.
6. Hence, the time taken is about 166 s.

Teacher's Note:
a) The disc is heated like a body charging through a thermal resistance, so the time follows an exponential law.
b) Heat capacity of the cylinder itself is ignored, as the question states.
c) Use ln 2 = 0.693.

14. An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross-section 8.0 × 10^-3 m². Initially the gas is at 300 K and occupies a volume of 2.4 × 10^-3 m³ and the spring is in its relaxed state. The gas is heated by a small heater until the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas. The force constant of the spring is 8000 N/m, and the atmospheric pressure is 1.0 × 10⁵ N/m². The cylinder and the piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and the cylinder. Neglect and heat loss through the lead wires of the heater. The heat capacity of the heater coil is negligible.

[Figure: A horizontal cylinder closed at the left end, with a heater coil connected to two terminals inside the gas on the left. A movable piston separates the gas from a spring on the right; the spring is attached to a fixed wall at the far right.]

Answer:
1. Initially the spring is relaxed, so the gas pressure p₁ = p₀ = 1.0 × 10⁵ N/m² and V₁ = 2.4 × 10⁻³ m³.
2. The piston moves out by x = 0.1 m, so the volume increases by Ax = 8.0 × 10⁻³ × 0.1 = 0.8 × 10⁻³ m³ and V₂ = 3.2 × 10⁻³ m³.
3. Final pressure p₂ = p₀ + kx/A = 1.0 × 10⁵ + (8000 × 0.1)/(8.0 × 10⁻³) = 2.0 × 10⁵ N/m².
4. From p₁V₁/T₁ = p₂V₂/T₂: T₂ = 300 × (2.0 × 10⁵ × 3.2 × 10⁻³)/(1.0 × 10⁵ × 2.4 × 10⁻³) = 800 K.
5. Hence, the final temperature of the gas is 800 K.

Teacher's Note:
a) The spring force adds to the atmospheric pressure on the outside of the piston.
b) The mass of gas is fixed, so the combined gas law applies.

15. Figure shows a process ABCA performed on n moles of an ideal gas.
Find the net heat given to the system during the process.

Answer:
1. In the V-T graph, A to B is at constant volume V₁ (T₁ to T₂), B to C is at constant temperature T₂ (V₁ to V₂), and C to A is a straight line through the origin, so V is proportional to T and the pressure is constant.
2. For a complete cycle ΔU = 0, so net heat Q = net work W.
3. W_AB = 0 (isochoric).
4. W_BC = nRT₂ ln(V₂/V₁) (isothermal expansion).
5. W_CA = p(V₁ - V₂) = nR(T₁ - T₂) (isobaric compression), since pV₁ = nRT₁ and pV₂ = nRT₂.
6. Net heat Q = nR[T₂ ln \( \frac{V_2}{V_1} \) - (T₂ - T₁)].

Teacher's Note:
a) Identify the type of each process from the shape of the V-T graph.
b) A straight line through the origin on a V-T graph means constant pressure.
c) In a cycle, the heat absorbed equals the work done.

16. Figure shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height h₀ and pressure 2p₀ where p₀ is the atmospheric pressure. There is a hole in the wall of the tank at a depth h₁ below the top from which water comes out. A long vertical tube is connected as shown.
(a) Find the height h₂ of the water in the long tube above the top initially.
(b) Find the speed with which water comes out of the hole.
(c) Find the height of the water in the long tube above the top when the water stops coming out of the hole.

[Figure: A large closed cylindrical tank containing water; the air space at the top has height h0 and pressure 2p0. A long vertical tube on the left side of the tank rises above the top of the tank, with the height h2 of water in it marked above the tank top. On the right, a hole in the tank wall, at depth h1 below the top, has a small downward-sloping spout from which water comes out.]

Answer:
1. (a) The pressure at the water surface inside the tank is 2p₀. The water surface in the long tube is a height (h₂ + h₀) above it and is open to the atmosphere, so 2p₀ = p₀ + ρg(h₂ + h₀).
2. So h₂ = \( \frac{p_0}{\rho g} \) - h₀.
3. (b) The hole is at a depth (h₁ - h₀) below the water surface in the tank. By Bernoulli's equation, 2p₀ + ρg(h₁ - h₀) = p₀ + (1/2)ρv².
4. So v = \( \left[ \frac{2}{\rho} \left( p_0 + \rho g (h_1 - h_0) \right) \right]^{1/2} \).
5. (c) Water stops coming out when the pressure at the hole becomes p₀. Then the water in the long tube stands at the level of the hole, which is h₁ below the top.
6. Hence, the height of water in the tube above the top is -h₁, that is, h₁ below the top.

Teacher's Note:
a) In part (a) the air pressure over the water in the tank is the pressure at the base of the tube column.
b) Note that the depth of the hole below the water surface is h₁ - h₀, not h₁.
c) A negative height means that the level is below the top of the tank.

17. The three rods shown in figure have identical geometrical dimensions. Heat flows from the hot end at a rate of 40 W in the arrangement: (a) Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c) Thermal conductivities of aluminium and copper are 200 W/m°C and 400 W/m-° C respectively.

Answer:
1. Let A/L be the common value of area/length for one rod. A rod with conductivity K has conductance K(A/L).
2. Arrangement (a): Al, Cu and Al in series between 0°C and 100°C. 1/G = (1/200 + 1/400 + 1/200)(L/A) = (5/400)(L/A).
3. Heat flow in (a) = ΔT/(1/G) = 100 × (400/5)(A/L) = 8000 A/L = 40 W, so A/L = 0.005 m.
4. Arrangement (b): one Al rod in series with the Cu and Al rods in parallel. The parallel pair has conductance (400 + 200)(A/L) = 600 A/L.
5. 1/G = (1/200 + 1/600)(L/A) = (1/150)(L/A), so heat flow = 100 × 150 × 0.005 = 75 W.
6. Arrangement (c): Al, Cu and Al rods in parallel. G = (200 + 400 + 200)(A/L) = 800 A/L, so heat flow = 100 × 800 × 0.005 = 400 W.
7. Hence, the rates of heat flow are 75 W in (b) and 400 W in (c).

Teacher's Note:
a) Rods in series add their thermal resistances; rods in parallel add their conductances.
b) Use the 40 W of arrangement (a) to find the common geometry factor.

18. A hot body placed in a surrounding of temperature θ₀ obeys Newton's law of cooling \( \frac{d\theta}{dt} \) = -k(θ - θ₀).
Its temperature at t = 0 is θ₁. The specific heat capacity of the body is s and its mass is m. Find
(a) the maximum heat that the body can lose and
(b) the time starting from t = 0 in which it will lose 90% of this maximum heat.

Answer:
1. (a) The body can cool only down to the surrounding temperature θ₀. Maximum heat lost = ms(θ₁ - θ₀).
2. (b) Solving dθ/dt = -k(θ - θ₀) gives θ - θ₀ = (θ₁ - θ₀)e^-kt.
3. Heat lost in time t = ms(θ₁ - θ) = ms(θ₁ - θ₀)(1 - e^-kt).
4. Setting this equal to 0.9 ms(θ₁ - θ₀): 1 - e^-kt = 0.9, so e^-kt = 0.1.
5. Hence, t = \( \frac{\ln 10}{k} \).

Teacher's Note:
a) The temperature approaches θ₀ exponentially and never goes below it.
b) ln 10 = 2.303, so t = 2.303/k.

19. Four identical rods AB, CD, CF and DE are joined as shown in figure. The length, cross-sectional area and thermal conductivity of each rod are l, A and K respectively. The ends A, E and F are maintained at temperatures T₁, T₂ and T₃ respectively. Assuming no loss of heat to the atmosphere, find the temperature at B

Answer:
1. B is the mid-point of rod CD, so BC and BD each have length l/2 and conductance 2KA/l. Rod AB has conductance KA/l.
2. Rods CF and DE have conductance KA/l each. Since C has no heat loss, the heat through BC equals the heat through CF, so BC and CF are in series.
3. Conductance of B to F = 1/(1/2 + 1) = 2/3 (in units of KA/l). Similarly B to E = 2/3.
4. At B the heat entering equals the heat leaving: 1 × (T₁ - T_B) + (2/3)(T₃ - T_B) + (2/3)(T₂ - T_B) = 0.
5. T_B (7/3) = T₁ + (2/3)(T₂ + T₃).
6. Hence, T_B = \( \frac{3T_1 + 2(T_2 + T_3)}{7} \).

Teacher's Note:
a) Treat each rod as a thermal resistance and apply Kirchhoff's junction rule at B.
b) The junctions C and D carry no net heat to the surroundings, so the rods in line carry the same current.
c) If T₁ = T₂ = T₃, then T_B equals that common value, a quick check.