Question

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

• A. ${325}^{o}C$
• B. $325K$
• C. ${39}^{o}C$
• D. ${320}^{o}C$

Answer 1

Answer: (C)

${39}^{o}C$

Given : $T_H=-{13}^o$ C $\beta =5$

$\therefore$ $T_H=273-13=260$ K

Coefficient of performance $\beta =\frac{T_H}{T_H-T_L}$

$\therefore$ $5=\frac{260}{260-T_L}$ $⟹T_L=312$ K

Thus temperature of air $T_L=312-273={39}^{o}C$