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

In a Carnot engine,when ${T_2} = {0^o}C$ and ${T_1} = {200^o}C$,its efficiency is ${\eta _1}$. When ${T_1} = {0^o}C$ and ${T_2} = -{200^o}C$,its efficiency is ${\eta _2}$. What is the ratio ${\eta _1}/{\eta _2}$?

$A$ Carnot engine working between $300\,K$ and $600\,K$ has a work output of $800\,J$ per cycle. What is the amount of heat energy supplied to the engine from the source per cycle in $J/cycle$?

$A$ heat engine operating between temperatures $T_1$ and $T_2$ has an efficiency of $\frac{1}{6}$. When $T_2$ is lowered by $62 \ K$,its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ respectively are:

$300 \, cal$ of heat is given to a heat engine and it rejects $225 \, cal$ of heat. If the source temperature is $227^{\circ} C$,then the temperature of the sink will be . . . . . . $^{\circ} C$.

$A$ Carnot engine has an efficiency of $50\%$ when its source is at a temperature $327^{\circ}C$. The temperature of the sink is $.........^{\circ}C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module