Question

Calculate the least amount of work that must be done to freeze one gram of water at $0^{0}C$ by means of the refrigerator.The temperature of the surrounding is ${27}^{0}C$.How much heat is passed on the surrounding in this process? Latent heat of fusion $L=80cal/g$.

• A. $87.91cal$
• B. $97.91cal$
• C. $88.95cal$
• D. $89.95cal$

Answer 1

The correct option is A $87.91cal$

$L=80\frac{cal}{g}$

$m=1g$

$T_1=27°=300K$

$T_2=0°=273K$

Least work done $W=L\times m\times \frac{T_1}{T_2}$

=$80\times 1\times \frac{300}{273}$

=$87.912cal$