Answer

Water is flowing at a rate of 3.0 litre/min.
The geyser heats the water, raising the temperature from 27°C to 77°C.
Initial temperature, T₁ = 27°C
Final temperature, T₂ = 77°C
∴Rise in temperature, ΔT = T₂ – T₁
= 77 – 27 = 50°C
Heat of combustion = 4 × 10⁴ J/g
Specific heat of water, c = 4.2 J g^–1 °C^–1
Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, ΔQ = mc ΔT
= 3000 × 4.2 × 50
= 6.3 × 10⁵ J/min
∴ Rate of consumption = 6.3 × 10⁵ / (4 × 10⁴)  =  15.75 g/min.

12.2. What amount of heat must be supplied to 2.0 × 10^–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N₂ = 28; R = 8.3 J mol^–1 K^–1.)

Answer

Mass of nitrogen, m = 2.0 × 10^–2 kg = 20 g
Rise in temperature, ΔT = 45°C
Molecular mass of N₂, M = 28
Universal gas constant, R = 8.3 J mol^–1 K^–1
Number of moles, n = m/M
= (2 × 10^-2 × 10³) / 28
= 0.714
Molar specific heat at constant pressure for nitrogen, C_p = (7/2)R
= (7/2) × 8.3
= 29.05 J mol^-1 K^-1
The total amount of heat to be supplied is given by the relation:
ΔQ = nC_P ΔT
= 0.714 × 29.05 × 45
= 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J.

12.3. Explain why
(a) Two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Answer

(a) In thermal contact, heat flows from the body at higher temprature to the body at lower temprature till tempratures becomes equal. The final temprature can be the mean temprature (T₁ + T₂)/2only when thermal capicities of the two bodies are equal.

(b) This is bcause heat absorbed by a substance is directly proportional to the specific heat of the substance.

(c) During driving, the temprature of air inside the tyre increases due to moion. Accordingto Charle’s law, P ∝ T. Therefore, air pressure inside the tyre increases.

(d) This is because in a harbour town, the relative humidity is more than in a desert town. hence, the climate of a harbour town is without extremes of hot and cold.

12.4. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Answer

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder = P₁
Final pressure inside the cylinder = P₂
Initial volume inside the cylinder = V₁
Final volume inside the cylinder = V₂
Ratio of specific heats, γ = 1.4
For an adiabatic process, we have:
P₁V₁^γ = P₂V₂^γ
The final volume is compressed to half of its initial volume.
∴ V₂ = V₁/2
P₁V₁^γ = P₂(V₁/2)^γ
P₂/P₁ = V₁^γ / (V₁/2)^γ
= 2^γ = 2^1.4 = 2.639
Hence, the pressure increases by a factor of 2.639.

12.5. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

Answer

The work done (W) on the system while the gas changes from state A to state B is 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.
∴ ΔQ = 0
ΔW = –22.3 J (Since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = ΔU + ΔW
Where,