# Question-1

**A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 ×10**

^{4}J/g?**Solution:**

Volume of water heated = 3.0 litres/min

Rise in temperature, ΔT = 77 - 27 = 50 °C

Specific heat of water, s = 4.2 J g

^{-1}°C

^{-1}

Amount of heat used, ΔQ = m ×s ×T

= 3000 ×4.2 ×50 = 63 × 10

^{4}J/min

Heat of combustion = 4 ×10

^{4}J/g

Rate of combustion of fuel = = 15.75 g/min.

# Question-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 N_{2}= 28) (R = 8.3J mol-1 K^{-1}).**Solution:**

Mass of the gas m = 2 × 10

^{- 2}Kg = 20g

Rise in temperature dt = 45°C

Heat required ∆Q = ?

Heat supplied = No. of moles × C

_{p}× dt = n C

_{P}× dt

Number of moles = = mole

C

_{p}= R (for diatomic molecule)

Rise in temperature dt = 45°C = 45K

Heat supplied = × × 8.3 × 45 = 933.75 J.

# Question-3

**d. The climate of a harbour town is more temperate than that of a town in desert at the same latitude.**

**Solution:**

a.

**In thermal contact, heat flows from the body at higher temperature to the body at a lower temperature till temperatures become equal. If the thermal capacities of the two bodies are equal then the mean temperature can be (T**

_{1}+T

_{2})/2 only.

b. Yes. Because heat absorbed by a body is directly proportional to the specific heat capacity of the body.

c. According to Charles law P ∝ T. During driving, the temperature of the air inside the tyre increases due to motion. So the temperature of the air inside the tyre increases.

In harbour town the relative humidity is more than in a desert town. So harbour town is more temperate than desert town.

# Question-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? γ =1.4.**

**Solution:**

# Question-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 23.3 J is done on the system. If the gas is taken from state B, an amount of work equal 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).**

**Solution:**

The system undergoes a cyclic change as it goes from A to B and then back to A. In a cyclic change there is no change in the internal energy of the system, i.e. dU = 0.

From the first law of thermodynamics, we have

dQ = dU + dW

dQ = dW [∴ dU = 0]

The amount of heat absorbed by the system

dQ = 9.35 cal - 22.3 J

= (9.35 × 4.2 - 22.3) J

= 16.8 J

Since dQ = dW, hence net work done by the system is 16.8 J.

# Question-6

**Two cylinders A and B of equal capacity are connected to each other via a stopcock. 'A' contains a gas at standard temperature and pressure. 'B' is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened.**

Answer the following:

(a) What is the final pressure of the gas in A and B?

(b) What is the change in internal energy of the gas?

(c) What is the change in the temperature of the gas?

(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface.

Answer the following:

(a) What is the final pressure of the gas in A and B?

(b) What is the change in internal energy of the gas?

(c) What is the change in the temperature of the gas?

(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface.

**Solution:**

(a) Since the vessels are of equal capacity, the volume occupied by the gas is doubled when the stopcock is opened. According to Boyle's law, the pressure of the gas is reduced to half. Hence the final pressure of the gas in A and B will be 0.5 atm.

(b) Since the system is thermally insulated from the surroundings, no transfer of energy takes place between the system and the surroundings. Hence there will be no change in the internal energy of the gas.

(c) For the same reason, no heat energy flows into the system or out of it. Hence there will be no change in the temperature of the gas.

(d) No, since the process (called free expansion) is rapid and cannot be controlled. In due course, the gas does return to an equilibrium state which lies on its P-V-T surface.

# Question-7

**A steam engine delivers 5.4 × 10**

^{8 }J of work per minute and services 3.6 × 10^{9 }J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?**Solution:**

Useful work done per min (Output) = 5.4 × 10

^{8}J

Heat absorbed per min (input) = 3.6 × 10^{9} J

Efficiency = output / input = 5.4 × 10^{8} J/3.6 × 10^{9} J = 0.15 =15%

Heat energy wasted per minute = Heat absorbed per minute – useful work done per minute

= 3.6 × 10^{9} J - 5.4 × 10^{8} J = 3.06 × 10^{9} J.

# Question-8

**An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing?**

**Solution:**

Heat supplied ∆Q=100 W = 100 J/s

Useful work done, ∆W = 75 J/s

Increase in internal energy = ∆U =?

As ∆Q = ∆U + ∆W

∆U = ∆Q - ∆W = 100 – 75 = 25 J/s.

# Question-9

**Solution:**

Change in pressure is dP = EF = 5.0 – 2.0 = 3.0 = 3.0 × 10

^{5}N/m

^{2}

Change in volume dV = DF = 600 – 300 = 300 cc = 300 × 10^{-6} m^{3}

Work done by the gas from D to E to F = area of ADEF

W = (1/2) × DF × EF = (1/2) × (300 × 10^{-6}) × (3.0 × 10^{5}) = 45 J.

# Question-10

**Refrigerators are to maintain eatables kept inside at 9° C, if room temperature is 36° C. Calculate the coefficient of performance.**

**Solution:**

T1 = 36 + 273 = 309 K

T2 = 9 + 273 = 202 K

.