Two ideal Carnot engines operate in cascade ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the first Carnot engine,the efficiency is $\eta_{1} = 1 - \frac{T}{T_{1}}$. The work done is $W_{1} = Q_{H1} \eta_{1} = Q_{H1} \left(1 - \frac

Vedclass · Accepted Answer

$T=\frac{T_{1}+T_{2}}{2}$

Vedclass · Answer

(C) For the first Carnot engine,the efficiency is $\eta_{1} = 1 - \frac{T}{T_{1}}$. The work done is $W_{1} = Q_{H1} \eta_{1} = Q_{H1} \left(1 - \frac{T}{T_{1}}\right)$.For the second Carnot engine,the efficiency is $\eta_{2} = 1 - \frac{T_{2}}{T}$. The work done is $W_{2} = Q_{H2} \eta_{2} = Q_{L1} \left(1 - \frac{T_{2}}{T}\right)$.Since the heat rejected by the first engine is the heat absorbed by the second engine,$Q_{L1} = Q_{H2}$.From the first engine,$Q_{L1} = Q_{H1} \left(\frac{T}{T_{1}}\right)$.Given $W_{1} = W_{2}$,we have $Q_{H1} \left(1 - \frac{T}{T_{1}}\right) = Q_{H1} \left(\frac{T}{T_{1}}\right) \left(1 - \frac{T_{2}}{T}\right)$.Simplifying,$1 - \frac{T}{T_{1}} = \frac{T}{T_{1}} - \frac{T_{2}}{T_{1}}$.$1 + \frac{T_{2}}{T_{1}} = \frac{2T}{T_{1}}$.Multiplying by $T_{1}$,we get $T_{1} + T_{2} = 2T$,which implies $T = \frac{T_{1} + T_{2}}{2}$.

Similar Questions

$A$ reversible engine converts one-sixth of the heat supplied into work. When the temperature of the sink is reduced by $62^{\circ} C$,the efficiency of the engine is doubled. The temperatures of the source and sink are

Write two essential characteristics of an ideal heat engine.

Even a Carnot engine cannot give $100\%$ efficiency because we cannot

The efficiency of a Carnot heat engine:

$A$ Carnot engine whose sink is at $300 \, K$ has an efficiency of $50 \%$. By how much should the temperature of the source be increased so that the efficiency becomes $70 \%$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module