Question

A Carnot engine A working between temperatures 600 K and T (T \(<\) 600 K) and another Carnot engine B working between temperatures T (T \(>\) 400 K) and 400 K are connected in series. If the work done by both the engines is same, then T =

• A. 550 K
• B. 500 K
• C. 575 K
• D. 525 K

Hint:

For a Carnot engine operating between \(T_H\) (hot reservoir) and \(T_L\) (cold reservoir): Efficiency \( \eta = 1 - \frac{T_L}{T_H} = \frac{W}{Q_H} \), where \(Q_H\) is heat absorbed from hot reservoir. Also, \( \frac{Q_L}{Q_H} = \frac{T_L}{T_H} \), where \(Q_L\) is heat rejected to cold reservoir. Work done \( W = Q_H - Q_L = Q_H(1 - T_L/T_H) \). If two Carnot engines are in series, heat rejected by the first is absorbed by the second. If work done is equal for \(T_1 \to T \to T_2\), then \(Q_1(1-T/T_1) = Q_1(T/T_1)(1-T_2/T)\). This simplifies to \(1-T/T_1 = (T/T_1) - T_2/T_1 \implies T_1-T = T-T_2 \implies 2T = T_1+T_2 \implies T = (T_1+T_2)/2\). Here \(T_1=600K, T_2=400K\). So \(T=(600+400)/2 = 500K\).

Answer 1

The Correct Option is B

Carnot Engine A: Source temperature \( T_{H1} = 600 \) K.
Sink temperature \( T_{L1} = T \).
Heat absorbed from source \( Q_{H1} \).
Heat rejected to sink \( Q_{L1} \).
Work done by engine A, \( W_A = Q_{H1} - Q_{L1} \).
Efficiency \( \eta_A = 1 - \frac{T_{L1}}{T_{H1}} = 1 - \frac{T}{600} \).
Also \( W_A = \eta_A Q_{H1} = \left(1 - \frac{T}{600}\right) Q_{H1} \).
For a Carnot cycle, \( \frac{Q_{L1}}{Q_{H1}} = \frac{T_{L1}}{T_{H1}} = \frac{T}{600} \implies Q_{L1} = Q_{H1} \frac{T}{600} \).
So, \( W_A = Q_{H1} \left(1 - \frac{T}{600}\right) \).
Carnot Engine B: The heat rejected by engine A, \( Q_{L1} \), is the heat absorbed by engine B.
So, heat absorbed by B, \( Q_{H2} = Q_{L1} = Q_{H1} \frac{T}{600} \).
Source temperature for B, \( T_{H2} = T \).
Sink temperature for B, \( T_{L2} = 400 \) K.
Work done by engine B, \( W_B = Q_{H2} - Q_{L2} \).
Efficiency \( \eta_B = 1 - \frac{T_{L2}}{T_{H2}} = 1 - \frac{400}{T} \).
\( W_B = \eta_B Q_{H2} = \left(1 - \frac{400}{T}\right) Q_{H2} = \left(1 - \frac{400}{T}\right) Q_{H1} \frac{T}{600} \).
\[ W_B = Q_{H1} \left(\frac{T-400}{T}\right) \frac{T}{600} = Q_{H1} \frac{T-400}{600} \] Given work done by both engines is the same: \( W_A = W_B \).
\[ Q_{H1} \left(1 - \frac{T}{600}\right) = Q_{H1} \frac{T-400}{600} \] Assuming \( Q_{H1} \ne 0 \): \[ \frac{600-T}{600} = \frac{T-400}{600} \] \[ 600-T = T-400 \] \[ 600+400 = T+T \] \[ 1000 = 2T \] \[ T = \frac{1000}{2} = 500 \, \text{K} \] Check conditions: \( T<600 \) K (500<600, true).
\( T>400 \) K (500>400, true).
This matches option (2).
This is a standard result for two Carnot engines in series with equal work output: the intermediate temperature T is the arithmetic mean of the source and final sink temperatures if efficiencies were equal, but here it's the arithmetic mean \( T = (T_H+T_L)/2 \) for work.
If efficiencies are equal, \( T = \sqrt{T_H T_L} \).
Here it's equal work.