91影视

The ratio of the heat removed from the low-temperature area inside a refrigerator to the work done by the refrigerator to remove the heat is termed the coefficient of performance of the refrigerator (COP).

The expression for coefficient of performance is

\({\rm{COP}} = \frac{{{Q_{\rm{L}}}}}{W}\).

In this problem, heat removed from the low-temperature region is the latent heat released during the conversion of water into the ice cubes.

The temperature inside the freezer is \({T_{\rm{L}}} = 0^\circ {\rm{C}} = \left( {0 + 273} \right)K = 273\;{\rm{K}}\).

The coefficient of performance of the cooling unit of the refrigerator is COP = 6.0.

The electrical power input of the refrigerator is \(P = 1.2\;{\rm{kW}} = 1.2 \times {\rm{1}}{{\rm{0}}^3}\;{\rm{W}}\).

The time taken is \(t = 1.0\;{\rm{h}} = 36{\rm{00}}\;{\rm{s}}\).

The density of water is\(\rho = 1.0 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\).

Latent heat of fusion of ice is \({L_f} = 3.33 \times {10^5}\;{\rm{J/kg}}\).

Let the volume of water converted into ice be V.

When water converts into ice cubes at \(0^\circ {\rm{C}}\), it releases the heat. The amount of heat released per unit mass of water is called the latent heat.

\(\begin{aligned}{l}\frac{{{Q_{\rm{L}}}}}{m} &= {L_f}\\{Q_{\rm{L}}} &= m{L_f}\end{aligned}\)

Since mass is density time its volume, the above expression can be re-written as

\({Q_{\rm{L}}} = \rho V{L_f}\).

The electric power input of the refrigerator is

\(\begin{aligned}{c}P &= \frac{W}{t}\\W &= P \times t.\end{aligned}\)

So, the expression for coefficient of performance of refrigerator is

\(\begin{aligned}{c}{\rm{COP}} &= \frac{{{Q_{\rm{H}}}}}{W}\\{\rm{COP}} &= \frac{{\rho V{L_f}}}{{P \times t}}.\end{aligned}\)

The expression for calculating the volume of water is

\(V = \frac{{COP \times P \times t}}{{\rho {L_f}}}\).

Substitute the values in the above expression.

\(\begin{aligned}{c}V &= \frac{{\left( {6.0} \right) \times \left( {1.2 \times {\rm{1}}{{\rm{0}}^3}\;{\rm{W}}} \right) \times \left( {3600\;{\rm{s}}} \right)}}{{\left( {1.0 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right) \times \left( {3.33 \times {{10}^5}\;{\rm{J/kg}}} \right)}}\\ &= 7.8 \times {10^{ - 2}}\;{{\rm{m}}^{\rm{3}}}\\ &= \left( {7.8 \times {{10}^{ - 2}}\;{{\rm{m}}^{\rm{3}}}} \right)\; \times \left( {\frac{{1000\;{\rm{L}}}}{{1\;{{\rm{m}}^{\rm{3}}}}}} \right)\\ &= 78\;{\rm{L}}\end{aligned}\)

Thus, the amount of water converted into ice in 1.0 h by the freezer is 78 L.