Question

A refrigerator freezes water at $0^{\circ }\text{C}$ into $10\text{kg}$ ice at $0^{\circ }\text{C}$ in a time interval of $30\text{min}$. Assuming the room temperature to be ${22}^{\circ }\text{C}$, calculate the minimum amount of power needed to make $10\text{kg}$ of ice.
(Take latent heat for freezing as $80\text{kcal/kg}$)

• A. $250\text{W}$
• B. $150\text{W}$
• C. $100\text{W}$
• D. $300\text{W}$

Answer 1

The correct option is B $150\text{W}$

As power is directly related to work done, so the work done to be minimum and hence the power, it should be based on Carnot cycle.

The amount of heat that must be extracted from the water to convert it into ice, hence for refrigerator
$\Rightarrow Q_2=\left(10\text{kg}\right)\left(80\text{kcal/kg}\right)=800\text{kcal}$
Here, $T_1={22}^{\circ }\text{C}=295\text{K}$ and
$T_2=273\text{K}$
Thus, coefficient of performance for ideal refrigerator,
$\beta =\frac{T_2}{T_1-T_2}=\frac{273}{295-273}=\frac{273}{22}$
If $W$ is the amount of work that must be done to make $10\text{kg}$ of ice,
$\Rightarrow \beta =\frac{Q_2}{W}$
$\therefore W=\frac{Q_2}{\beta }=\frac{800}{\left(273/22\right)}\simeq 64.5\text{kcal}$
$\Rightarrow W=64.5\times {10}^3\times 4.186\text{J}\simeq 2.7\times {10}^5\text{J}$
Since this work must be done in $30\text{min}$, power required to run the refrigerator is given by,
$\Rightarrow P=\frac{W}{t}$
$\Rightarrow P=\frac{2.7\times {10}^5\text{J}}{(30\times 60)\text{s}}=150\text{W}$