• question_answer 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) $325{}^\circ C$

 [b]Given $\beta =5$ We have, $\beta =\frac{{{T}_{2}}}{{{T}_{1}}-{{T}_{2}}}$ where ${{T}_{1}}=$ temp. of surrounding, ${{T}_{2}}=$ temp. of cold body $=-{{13}^{0}}C=(-13+273)K=260K$ or         $5=\frac{260}{{{T}_{1}}-260}$ $5{{T}_{1}}=260+1300$ or ${{T}_{1}}=312\,K={{39}^{o}}C$.