Question

How much energy in watt hour may be required to convert $2$ kg of water into ice at $0^{\circ}C$, assuming that the refrigerator is ideal? Given temperature of freezer is $-15^{\circ}C$, room temperature is$25^{\circ}C$ and initial temperature of water is $25^{\circ}C$.

Answer 1

Here $T_{1} = 273 + 25 = 298 K$ and $T_{2} = 273 - 15 = 258 K$
Specific heat of water, $C = 4.2 \times 10^{3} J/kg K$
Latent heat of fusion of ice, $L= 3.36 \times 10^{5} J/kg$
The amount of heat required to transform water of $25^{\circ}C$ into ice of $0^{\circ}C$
$Q_{2} = mC\Delta T +mL$
$ = 2 \times 4.2 \times 10^{3} \times \left(25 -0\right) + 2\times 3.36 \times 10^{5}$
$ = 2.1 \times 10^{5} + 6.72 \times 10^{5}$
$ = 8.82 \times 10^{5}J$
Heat rejected to the surroundings;
we have $\frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}}$
$\therefore Q_{1} = Q_{2} \left(\frac{T_{1}}{T_{2}}\right)$
$= 8.82 \times10^{5} \left(\frac{298}{258}\right) = 10.19 \times10^{5} J$
Energy supplied to convert water into ice,
$W = Q_{1} - Q_{2}$
$ = 10.15 \times 10^{5} -8 .8 2 \times 10^{5}$
$ = 1.33 \times 10^{5} J$
$= \frac{1.33 \times10^{5}}{3600} = 36.96$ Wh