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

Hint: A refrigerator is an anticlockwise operating Carnot cycle. The coefficient of performance defined for a refrigerator is similar to the efficiency in Carnot cycle that is ratio of output to input. But in a refrigerator, work is done on the system. Thus, the input is work and it is always positive. Unlike Carnot cycle, where work is done by the system and hence it is negative. The output is work in case of a Carnot cycle. Latent heat refers to the hidden heat that converts ice to water without change in temperature, hence it appears that there is no heat involved.

Formula used:
$\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$

Complete answer:
For a refrigerator,
$\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$------(1)
Where,
\[{Q_1}\] is the work done to convert water to ice
\[Q_2\]iIs the heat release to the surrounding
${T_1}$ is the temperature of water\[ = 0^\circ C\]$ = 273K$
${T_2}$is surrounding temperature\[ = 27^\circ C\]$ = 300K$

Now, we know that, the amount of heat required is directly proportional to the latent heat and can be calculated using the formula:
\[{Q_1} = mL\]
where L is the latent heat of ice and m is mass of water converted to ice

It is given that latent heat of ice is=L=\[80{\text{ }}cal/gm\]
Thus,
\[{Q_1} = 100\left( {80} \right) = 8000{\text{ }}cal\]

Putting values in equation 1
$\dfrac{{8000}}{{{Q_2}}} = \dfrac{{273}}{{300}}$
${Q_2} = 8791$cal

Thus, the heat released to the surrounding is $8791$cal.

Note: A refrigerator accepts heat from a lower temperature and rejects heat to higher temperature. Thus, a refrigerator (used in our houses) is used to further decrease the temperature of substances (food items). Before proceeding to the calculations, convert temperature into kelvin.