Question

A Carnot cycle consists of a cycle of 4 processes, in the order:

1) isothermal expansion at T(hot)=90oC,

2) adiabatic expansion to T(cold)=30oC,

3) isothermal contraction at T(cold),

4) adiabatic contraction to the original state.

The gas in the system can be treated as a monotonic ideal gas

Part A. After one complete cycle, what is the change in thermal energy of the gas in the system?

Part B. After a complete cycle what is the change in entropy? (The answer to this part is 0 J)

Part C.In this cycle heat can be transferred during which parts of the cycle? (The answer to this part is leg 1 and 3 only)

Part D. What is Q3/Q1, the ratio of heat transferred in leg 3 divided by heat transferred in leg 1 of the cycle?

Part E. What is the Work in terms of Q1 and Q3? (Hint: The work does not depend on Q1 and Q3)

Part F. What is the efficiency of the cycle?

Answer 1

Part A

change in thermal energy is proportional to change in temperature of gas i.e., T₂ - T₁

But for one complete cycle, starting temperature = end temperature or T₁ = T₂

Hence T₂ - T₁ = 0 ; which gives change in thermal energy = 0

Part B

Entropy is a state function and does not depend on path

Let S₁ = entropy at start of process, S₂ = entropy at end of process

Change in entropy = S₂ - S₁

But for one complete cycle of carnot cycle, end point is same as start point

Since the state is same, S₁ = S₂

Hence S₁ - S₂ = 0 ; which gives change in entropy = 0

Part C

Adiabatic Process means complete insulation of system and no heat nor mass is exchanged between system and surrounding. Hence heat cannot be transferred in 2^nd and 4^th part of cycle.

In isothermal process, mass cannot be exchanged between system and surrounding but heat flow has no restriction. Hence heat can be transferred during isothermal process i.e., during 1^st and 3^rd part of cycle.

Part D

Q₃/Q₁ = -1

This is because Q₃ is equal and opposite to Q₁

Part E

From 1^st law of thermodynamics, Delta U = Q - W

Where Delta U = net change in internal energy, Q = heat energy, W = work done

For isothermal process, Delta U = 0

Hence Q = W ; W₁ = Q₁ and W₃ = Q₃

Part F

Efficiency = 1 - (T₂ / T₁) , where T₂ = 30^oC and T₁ = 90^oC

efficiency = 1 - (30^oC / 90^oC)

efiiciency = 1 - 0.33

efficiency = 0.67