Question

A carnot engine takes 300 calories of heat at 500 K and rejects 150 calories of heat to the sink. The temperature of the sink is

• A. 1000 K
• B. 750 K
• C. 500 K
• D. 250 K

Answer 1

Answer: (D)

250 K

Answer 2

The efficiency of a carnot engine is given as
$\eta = 1 - \frac{T_{2}}{T_{1}} = 1 -\frac{Q_{2}}{Q_{1}}$
Where, $T_1 =$ source temperature, $T_2 =$ sink temperature
$Q_1 =$ heat extracted from source and $Q_2 =$ heat given to sink.
Here, $T_1 = 500 \,K, Q_1 = 300 \,cal$ and $Q_2 = 150 \,cal$
Now putting these values in above equation,
$1 - \frac{T_{2}}{500} = 1 - \frac{150}{300}$
$\Rightarrow T_{2} = 500\times \frac{150}{300} = 250\, K$
Hence, the temperature of sink is $250\, K$.