Question

A refrigerator placed in a room at 300 K has inside temperature at 264K. How many calories of heat shall be delivered to the room for each 1 kcal of energy consumed by the refrigerator ideally

• A. 7.2 kcal
• B. 7.83 kcal
• C. 8.33 kcal
• D. 8.94 kcal

Answer 1

The correct option is C 8.33 kcal

Since the refrigerator is reversible,
$\frac{Q_2}{Q_1}=\frac{T_2}{T_1}=\frac{264}{300}$
where, $T_1$ = room temperature, $Q_1$ =heat released
$T_2$ = refrigerator temperature, $Q_2$ = heat absorbed
From first law,
$Q_1-Q_2$ =1 Kcal
Thus solving above equations we get, heat released= 8.33 Kcal