Question

In a Carnot heat engine, the temperature of source and sink are ${827}^{\circ }\text{C}$ and ${27}^{\circ }\text{C}$. If the engine consumes $33\text{kJ}$ per cycle, then the heat rejected to the sink per cycle is

• A. $11\text{kJ}$
• B. $9\text{kJ}$
• C. $24\text{kJ}$
• D. $10\text{kJ}$

Answer 1

The correct option is B $9\text{kJ}$

Temperature of source, $T_1={827}^{\circ }\text{C}=1100\text{K}$
Temperature of sink, $\left(T_2\right)={27}^{\circ }\text{C}=300\text{K}$
Heat absorbed, $\left(Q_1\right)=33\text{kJ}$
Efficiency of the heat engine will be,
$\eta =1-\frac{T_2}{T_1}=1-\frac{300}{1100}$
$\therefore \eta =\frac{8}{11}$
From the fundamental definition of efficiency for any heat engine,
$\eta =\frac{W}{Q_1}$
$\Rightarrow W=\eta Q_1$
$W=\frac{8}{11}\times 33\times {10}^3\text{J}=24\text{kJ}$
Heat rejected to the sink,
$\Rightarrow Q_2=Q_1-W$
$\therefore Q_2=33\text{kJ}-24\text{kJ}=9\text{kJ}$

ALTERNATIVE:
For a Carnot heat engine,
$\frac{Q_1}{Q_2}=\frac{T_1}{T_2}$
$\Rightarrow Q_2=Q_1\times \frac{T_2}{T_1}$
$\therefore Q_2=33\times {10}^3\times \frac{3}{11}=9\text{kJ}$