At Saralstudy, we are providing you with the solution of Class 11 Physics Thermodynamics according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. We are trying our best to give you detailed answers to all the questions of all the topics of Class 11 Physics Thermodynamics so that you can prepare for the exam according to your own pace and your speed.
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 = = 15.75 g/min
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
= 2.0 x 10-2 x 103 / 28 = 0.714
Molar specific heat at constant pressure for nitrogen, CP = 7/2R
= 7/2 x 8.3
= 29.05 J mol-1 K-1
The total amount of heat to be supplied is given by the relation:
ΔQ = nCP ΔT
= 0.714 × 29.05 × 45
= 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J.
(a) When two bodies at different temperatures T1 and T2 are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T1 + T2)/2 only when the thermal capacities of both the bodies are equal.
(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.
(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.
(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.
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,
ΔU = Change in the internal energy of the gas
∴ ΔU = ΔQ - ΔW = - (- 22.3 J)
ΔU = + 22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is:
ΔQ = 9.35 cal = 9.35 x 4.19 = 39.1765 J
Heat absorbed, ΔQ = ΔU + ΔQ
∴ΔW = ΔQ - ΔU = 39.1765 - 22.3 = 16.8765 J
Therefore, 16.88 J of work is done by the system.
Work done by the steam engine per minute, W = 5.4 × 108 J
Heat supplied from the boiler, H = 3.6 × 109 J
Efficiency of the engine = Output energy / Input energy
∴ n = W / H
= 5.4 × 108 / 3.6 × 109
Hence, the percentage efficiency of the engine is 15 %.
Amount of heat wasted = 3.6 × 109 - 5.4 × 108
= 30.6 × 108 = 3.06 × 109 J
Therefore, the amount of heat wasted per minute is 3.06 × 109 J.
Heat is supplied to the system at a rate of 100 W.
∴Heat supplied, Q = 100 J/s
The system performs at a rate of 75 J/s.
∴Work done, W = 75 J/s
From the first law of thermodynamics,we have:
Q = U + W
Where, U = Internal energy
∴U = Q - W
= 100 - 75
= 25 J/s
= 25 W
Therefore, the internal energy of the given electric heater increases at a rate of 25 W.
Temperature inside the refrigerator, T1 = 9°C = 282 K
Room temperature, T2 = 36°C = 309 K
Coefficient of performance = T1 / T2 - T1
= 282 / 309 - 282
= 282 / 27
= 10.44
Therefore, the coefficient of performance of the given refrigerator is 10.44.