Question

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

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

Answer 1

Answer: (B)

Given: $Q_1 = 1000\,J$
$Q_2 = 600\,J$
$T_1=127^°C = 400\,K$
$T_2= ?$
$\eta=?$
Efficiency of carnot engine,
$\eta=\frac{W}{Q_{1}}\times100\%$
or, $\eta=\frac{Q_{2}-Q_{1}}{Q_{1}}\times 100\%$
$\eta=\frac{1000-600}{1000}\times 100\%$
$\eta=40\%$
Now, for carnot cycle $\frac{Q_{2}}{Q_{1}}=\frac{T_{2}}{T_{1}}$
$\frac{600}{1000}=\frac{T_{2}}{400}$
$T_{2}=\frac{600\times400}{1000}=240\,K=240-273$
$\therefore T_{2}=-33^{°}C$