Question

A heat engine is working between $260K$ and $300K$. It takes $500$ Cal. of heat from sink. Heat rejected to the source at higher temperature for this refrigerator is

• A. $400$ Cal.
• B. $477$ Cal.
• C. $377$ Cal.
• D. $577$ Cal.

Answer 1

Answer: (D)

$577$ Cal.

Efficiency of the heat engine$\left(\eta \right)=\frac{Q_h-Q_c}{Q_h}=1-\frac{T_c}{T_h}$
$1-\frac{Q_c}{Q_h}=1-\frac{T_c}{T_h}$
$\frac{500}{Q_h}=\frac{260}{300}$
$Q_h=500\times \frac{300}{260}=576.92K$
Option D.