Question

A “Carnot” refrigerator (the reverse of a Carnot engine) absorbs heat from the freezer compartment at a temperature of -17°C and exhausts it into the room at 25°C.

(a) How much work would the refrigerator do to change 0.65 kg of water at 25°C into ice at -17°C.

(b) If the compressor output is 105 W and runs 25% of the time, how long will this take?

Short answer

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

(b) The time taken to do the work is\(32\;\min \).

Step-by-step solution

Heat removed from the water

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.

Given data

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 \(P = 105\,{\rm{W}}\).

Evaluation of heat rejected from the water

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.

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

(a) Evaluation of the work done by the refrigerator

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 \(5.03 \times {104}\;{\rm{J}}\).

(b) Evaluation of the time consumed by using compressor power

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 \(32\;\min \).