Question

A heat engine operating between ${227}^{\circ }C$ and ${27}^{\circ }C$ absorbs 2 kcal of heat from the ${227}^{\circ }C$ reservoir reversibly per cycle. The amount of work done in one cycle is ?

• A. 0.4 kcal
• B. 0.8 kcal
• C. 4 kcal
• D. 8 kcal

Answer 1

The correct option is B 0.8 kcal

We have

$T_1=27+273=298K$

$T_2=227+273=500K$

$\eta =\frac{T_2-T_1}{T_2}$

$\eta =\frac{500-300}{500}=\frac{2}{5}$

$\eta =\frac{W_{by}}{q_{source}}\Rightarrow W_{by}=\left(n_C\right)\left(q_{source}\right)$

$=\left[\frac{2}{5}\right]\left(2kcal\right)$

$=0.8kcal$

Hence option B is correct.