Question

How much energy in a watt hour may be required to convert \[2kg\] of water into ice at $ {0^ \circ }C$ , assuming that the refrigerator Is ideal? Take room temperature $ = {25^ \circ }C$ , which is also the initial temperature of water and temperature of freezer is $ - {15^ \circ }C$
A. $7{\text{ }}Wh$
B. $17\;Wh$
C. $50{\text{ }}Wh$
D. $38{\text{ }}Wh$

Answer 1

Hint: To convert water into ice. Water must go through the phase change.The phase change is carried out by losing the latent heat of freezing. If we were going to evaporate the water in that case the latent heat of vaporization must be added.The process of adding or losing latent heat is carried out at constant temperature.

Complete step by step answer:
Given data, the mass of water is 2Kg.
Room temperature $ = {25^ \circ }C$
Temperature of freezer is $ - {15^ \circ }C$
Converting the temperature into Kelvin we get,
Room temperature $ = 25 + 273 = 298K$
Freezer temperature $ = - 15 + 273 = 258K$
Specific heat of water is $ {C_p} = 4.2kJ/kg$
Latent heat of ice is $ L = 336kJ/kg$
We need to remove the following amount of heat in order to form ice at $ {0^ \circ }C$ .
$ {Q_2} = m{C_p}\Delta T + mL$
Putting the values in equation we get,
$ {Q_2} = 2 \times 4.2 \times (25 - 0) + 2 \times 336 \\
\Rightarrow {Q_2} = 210 + 672 \\
\Rightarrow {Q_2} = 882\,kJ/kg \\ $
Now from the formula of heat pump we have
\[\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{T_1}}}{{{T_2}}} \\
\Rightarrow {Q_1} = \dfrac{{{T_1}}}{{{T_2}}} \times {Q_2} \\ \]
Substituting the values in the equation we will get,
${Q_1} = \dfrac{{298}}{{258}} \times 882 \\
\Rightarrow {Q_1} =1015\,kJ/kg \\ $
The external energy we have to supply v=can be found out using
$W = {Q_1} - {Q_2} \\
\Rightarrow W = (1015 - 882)kJ \\
\Rightarrow W =133\,kJ \\ $
We can multiply and divide by hour to get the rating in Watt hour
1 h =3600 s
$\dfrac{{133000}}{{3600}}Wh \\
\Rightarrow 37.96Wh \approx 38Wh \\ $
Hence, the correct answer is option D.

Note:The heat transferred or the external work done is given by the difference in the heat accepted and heat lost. For any phase changes we require latent heat to be transferred to the substance. Transfer of latent heat occurs at the constant temperature. These constant temperature processes are called isothermal processes.