Question

A Carnot engine is operating between a hot body and cold body maintained at temperature $T_1$ and $T_2$ respectively. Consider the following three cases : The temperature of the hot body is changed to $T_1 + \Delta T$ and cold body is at $T_2$ : The temperature of the hot body is at $T_1$ and cold body is changed to $T_2 + \Delta T$ : The temperature of the hot body is at $T_1$ and cold body is changed to $T_2 - \Delta T$

• A. The efficiency of the Carnot cycle is highest for case -I
• B. The efficiency of the Carnot cycle is highest for case -II
• C. The efficiency of the Carnot cycle is highest for case -III
• D. The efficiency of case -II is higher than case -III

Answer 1

Answer: (C)

The efficiency of the Carnot cycle is highest for case -III

Answer 2

We know that efficiency of Carnot engine is given by relation
$\eta=\frac{T_{1}-T_{2}}{T_{1}}$
Hence in
$\eta_{ I } =\frac{\left(T_{1}+\Delta T\right)-T_{2}}{T_{1}+\Delta T}$
$=\frac{\left(T_{1}-T_{2}\right)+\Delta T}{T_{1}+\Delta T}$
$\eta_{ II } =\frac{T_{1}-\left(T_{2}+\Delta T\right)}{T_{1}}$
$\eta_{ III } =\frac{T_{1}-\left(T_{2}-\Delta T\right)}{T_{1}}$
$=\frac{\left(T_{1}-T_{2}\right)+\Delta T}{T_{1}}$
So, from above efficiences we can conclude the order as
$\eta_{ III }>\eta_{ II }>\eta_{ I }$