Question

An ideal refrigerator has a freezer at a temperature of $-13^{\circ} C$. The coefficient of performance of the engine is $5$. The temperature of the air (to which heat is rejected) is

• A. $320^{\circ} C$
• B. $39^{\circ} C$
• C. $325\, K$
• D. $325^{\circ} C$

Answer 1

Answer: (B)

$39^{\circ} C$

Answer 2

Given: $T _{ H }=-13^{\circ} C\,\, \beta=5$
$\therefore T _{ H }=273-13=260\, K$
Coefficient of performance
$\beta=\frac{ T _{ H }}{ T _{ H }- T _{ L }}$
$\therefore 5=\frac{260}{260- T _{ L }}$
$\Rightarrow T _{ L }=312\, K$
Thus temperature of air
$T_{L}=312-273=39^{\circ} C$