91Ó°ÊÓ

Lower temperature$T_o=0°C=273K$

Higher temperature $T_K=25°C=298K$

Mass of liquid water, m$=10Kg$.

Formula used :

The coefficient of performance of refrigerator is calculated as ${\left(CO{P}_{Carnot}\right)}_{ref}=\frac{T_o}{T_K-T_o}=\frac{Heatremoved}{Heatexhausted-Heatremoved}$

Here,

$T_o$is the lower temperature

$T_K$is the higher temperature

Calculation :

Heat required to remove from the water to convert it into ice is calculated as$Q_{removed}=m{L}_{ice}$

Here,

$L_{ice}=334kJ/kg$is latent heat of fusion

Plugging the values in the above equation $$\begin{gathered} Q_{removed}=10×334 \\ =3340kJ \end{gathered}$$

Now the heat exhausted into the room is calculated as $\frac{T_o}{T_k-T_o}=\frac{Heatremoved}{heatexhausted-Heatremoved}$

Plugging the values in the above equation$$\begin{gathered} \frac{273}{298-273}=\frac{3340}{Q_{exhaust}-3340} \\ {Q}_{exhaust}=3645.8kJ \end{gathered}$$

Conclusion :

Heat exhausted into the room is$3645.8kJ$.

Formula used :

Energy supplied to the refrigerator is calculated as $Q_{Supplied}=Q_{exhaust}-Q_{remove}$

Calculation :

Energy supplied to the refrigerator is calculated as $Q_{Supplied}=Q_{exhaust}-Q_{remove}$

Plugging the values in the above equation $$\begin{gathered} Q_{Supplied}=3645.8-3340 \\ =3.5.8kJ \end{gathered}$$

Conclusion :

Energy supplied to the refrigerator is$305.8kJ$.