Question

A Carnot engine absorbs $1000J$ of heat energy from a reservoir at ${127}^{o}C$ and rejects $600J$ of heat energy during each cycle. The efficiency of engine and temperature of sink will be

• A. $20$% and $-{43}^{o}C$
• B. $40$% and $-{33}^{o}C$
• C. $50$% and $-{20}^{o}C$
• D. $70$% and $-{10}^{o}C$

Answer 1

Answer: (B)

$40$% and $-{33}^{o}C$

$\text{Given:-}$ $Q_1=1000J$

$Q_2=600J$

$T_1={127}^{\circ }C=400K$

$\text{Solution:-}$

Efficiency of carnot engine,

$\eta =\frac{W}{Q_1}\times 100$ or

$\eta =\frac{Q_2-Q_1}{Q_1}\times 100$

$\eta =\frac{1000-600}{1000}\times 100$

$\eta =40$%

Now, for carnot cycle $\frac{Q_2}{Q_2}=\frac{T_2}{T_1}$

$\frac{600}{1000}=\frac{T_2}{400}$

$T_2=\frac{600\times 400}{1000}$

$=240K$

$=240-273$

$\therefore T_2=-{33}^{\circ }C$

$\text{Hence the correct option is B}$