91影视

In this problem, there are three parts in which the heat rejected is evaluated. First, the cooling of water up to the freezing point; second, freezing the liquid water; and last, cooling the ice.

The temperature of the freezer is \({T_{\rm{1}}} = - 17\circ {\rm{C}}\).

The temperature of the room is \({T_{\rm{2}}} = 25\circ {\rm{C}}\).

The mass is \(m = 0.65\,{\rm{kg}}\).

The output power is \(P = 105\,{\rm{W}}\).

The relation to find the heat rejected is given by:

\(\begin{array}{l}Q = mc\Delta T + mL + mc'\Delta T'\\Q = m\left( {c\Delta T + L + c'\Delta T'} \right)\end{array}\)

Here, \(c\)and \(c'\) are the specific heat for water and ice, \(\Delta T\) and \(\Delta T'\) are the change in temperature for water and ice, and Lis the latent heat of water.

Substitute the values in the above expression.

\(\begin{array}{l}Q = \left( {0.65\;{\rm{kg}}} \right)\left( {\left( {4186\;{\rm{J/kg}} \cdot \circ {\rm{C}}} \right)\left( {25\circ {\rm{C}}} \right) + \left( {3.33 \times {{10}5}\;{\rm{J/kg}}} \right) + \left( {2100\;{\rm{J/kg}} \cdot \circ {\rm{C}}} \right)\left( {17\circ {\rm{C}}} \right)} \right)\\Q = 3.07 \times {105}\;{\rm{J}}\end{array}\)

The relation of efficiency is given by:

\(\begin{array}{c}\eta = \frac{W}{{W + Q}}\\1 - \frac{{{T_{\rm{1}}}}}{{{T_{\rm{2}}}}} = \frac{W}{{W + Q}}\\W = Q\left( {\frac{{{T_2}}}{{{T_1}}} - 1} \right)\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{l}W = \left( {3.07 \times {{10}5}\;{\rm{J}}} \right)\left( {\frac{{\left( {25\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}{{\left( { - 17\circ {\rm{C}} + 273} \right)\;{\rm{K}}}} - 1} \right)\\W = 5.03 \times {104}\;{\rm{J}}\end{array}\)

Thus, the work done by the refrigerator is \(5.03 \times {104}\;{\rm{J}}\).

The relation to find the time is given by:

\(t = \frac{W}{P}\)

Substitute the values in the above expression.

\(\begin{array}{l}t = \left( {\frac{{5.03 \times {{10}4}\;{\rm{J}}}}{{105\,{\rm{W}}\left( {0.25} \right)}}} \right)\\t = 1916.1\;{\rm{s}}\\t \approx 32\;\min \end{array}\)

Thus, the time taken to do the work is \(32\;\min \).