If minimum possible work is done by a refrigerator in converting 100g of water at  $0^{o}C$ to ice, how much heat (in cal) is released to surroundings at temperature  ${27}^{o}C$ to the nearest integer ____? (Take, latent heat of ice = 80 cal/g)

To convert 100g of water at  $0^{o}C$ to ice at  $0^{o}C$ , heat absorbed by refrigerator,  
$Q_1=mL=100\times 80=8000cal$

$W=$ workdone or energy supplied to refrigerator
 
From figure, we have
 $W+Q_1=Q_2\Rightarrow W=Q_2-Q_1$
Now, coefficient of performance of refrigerator is 
$\beta =\frac{Q_1}{W}=\frac{Q_1}{Q_2-Q_1}=\frac{T_1}{T_2-T_1}$ 
Here,  $T_1=0^{o}C=273K$
         $T_2={27}^{o}C=300K$  
So, $\frac{Q_1}{Q_2-Q_1}=\frac{273}{300-273}$

$$\begin{gathered} 27Q_1=(Q_2-Q_1)273 \\ 300Q_1=Q_2\times 273 \\ {Q}_2=300\times \frac{8000}{273} \\ =\frac{2400000}{273} =8791.2cal \simeq 8791cal \end{gathered}$$