Thermodynamics

1. 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
2. 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,
   
   Molar specific heat at constant pressure for nitrogen,
   
   
   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.
3. (a) When two bodies at different temperatures T₁and 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.
4. 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:
   
   The final volume is compressed to half of its initial volume.
   ∴
   
   Hence, the pressure increases by a factor of 2.639.
    
5. 
   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 × 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.
6. Answer:
   (a) 0.5 atm
   (b) Zero
   (c) Zero
   (d) No
   Explanation:
   (a) The volume available to the gas is doubled as soon as the stopcock between cylinders A and B is opened. Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5 atm.
   (b) The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change.
   (c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.
   (d) The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non-equilibrium states, they do not lie on the P-V-T surface of the system.
    
7. 
   Work done by the steam engine per minute, W= 5.4 × 10⁸J
   Heat supplied from the boiler, H= 3.6 × 10⁹J
   Efficiency of the engine =
   
   Hence, the percentage efficiency of the engine is 15 %.
   Amount of heat wasted = 3.6 × 10⁹– 5.4 × 10⁸
   = 30.6 × 10⁸= 3.06 × 10⁹J
   Therefore, the amount of heat wasted per minute is 3.06 × 10⁹J.
8. 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.
9. Total work done by the gas from D to E to F = Area of ΔDEF
   Area of ΔDEF =
   Where,
   DF = Change in pressure
   = 600 N/m² – 300 N/m²
   = 300 N/m²
   FE = Change in volume
   = 5.0 m³ – 2.0 m³
   = 3.0 m³
   Area of ΔDEF = = 450 J
   Therefore, the total work done by the gas from D to E to F is 450 J.
10. Temperature inside the refrigerator, T₁ = 9°C = 282 K
    Room temperature, T₂ = 36°C = 309 K
    Coefficient of performance =
    
    Therefore, the coefficient of performance of the given refrigerator is 10.44.