Question

If the minimum possible work is done by a refrigerator in converting $100\text{grams}$ of water at $0{}^{\circ }\text{C}$ to ice, how much heat (in calories) is released to the surroundings at temperature $27{}^{\circ }\text{C}$ (Latent heat of ice = 80 Cal/gram) to the nearest integer?

Answer 1

Heat absorbed is given by,
$Q_2=mL=80\times 100=8000\text{Cal}$

Temperature of ice, $T_2=0{}^{\circ }\text{C}=273\text{K}$

Temperature of surrounding, $T_1=273+27=300\text{K}$

Coefficient of performance
$\alpha =\frac{W}{Q_2}=\frac{Q_1-Q_2}{Q_2}$

As minimum work is done, this can be considered as ideal refrigerator,

$\Rightarrow \frac{Q_1-Q_2}{Q_2}=\frac{T_1-T_2}{T_2}$

$\Rightarrow \frac{Q_1-8000}{8000}=\frac{27}{273}$

$\Rightarrow Q_1=8791\text{Cal}$

Hence, $8791$ is the correct answer.