## Related questions with answers

A freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80 kg of water at $25.0^{\circ} \mathrm{C}$ to 1.80 kg of ice at $-5.0^{\circ} C$ in one hour. (a) What amount of heat must be removed from the water at $25.0^{\circ} C$ to convert it to ice at $-5.0^{\circ} C$? (b) How much electrical energy is consumed by the freezer during this hour? (c) How much wasted heat is delivered to the room in which the freezer sits?

Solution

Verified## Given

We are given a coefficient of performance $K$ = 2.40 for a freezer. A mass of water $m$ =1.80 kg at temperature $T_{w} = 25.0^{\circ} \text{C}$ convert with the same mass to ice and reach the temperature $T_{ice} =-5.0^{\circ} \text{C}$.

## Solution

**(a)** We want to calculate the removed heat from the water. In this problem, the water has three processes happened. First the water-cooled from temperature $25.0^{\circ}\text{C}$ to $0.0 ^{\circ} \text{C}$ so the heat in this process will be given by

$\begin{aligned} Q &= mc_{w}\Delta T = mc_{w} (0.0^{\circ}\text{C}- 25.0^{\circ} \text{C}) \tag{1} \end{aligned}$

Where $c_{w}$ is the specific heat capacity of water and equals 4190 $\mathrm{J/kg\cdot C^{o}}$ (See Table 17.3).

In the second process, the water change from the liquid phase to the solid phase, and the heat will be given by equation 17.20

$Q = -mL_{f} \tag{2}$

Where $L_{f}$ is the heat fusion of water and equals 3.34 $\times 10^{5}$ J/kg. (See Table 17.4)

The third process, the ice temperature changes from $0.0^{\circ} \text{C}$ to $-5.0^{\circ} \text{C}$ and the heat in this process will be given by

$\begin{aligned} Q &= mc_{ice}\Delta T = mc_{ice} (-5.0^{\circ}\text{C}- 0.0^{\circ} \text{C}) \tag{3} \end{aligned}$

Where $c_{ice}$ is the specific heat capacity of the ice and equals 2100 $\mathrm{J/kg\cdot C^{o}}$ (See table 17.3).

Now we can obtain the heat removed from the water by combine the three processes from equations (1), (2) and (3)

$\begin{aligned} Q &= mc_{w} (- 25.0^{\circ} \text{C}) -mL_{f} + mc_{ice} (-5.0^{\circ}\text{C}) \\ &= 1.80 \,\text{kg} \times 4190 \mathrm{~J/kg\cdot C^{o}} (- 25.0^{\circ} \text{C}) - 1.80 \,\text{kg} \times 3.34 \times 10^{5} \,\text{J/kg} + 1.80 \,\text{kg} \times 2100 \mathrm{~J/kg\cdot C^{o}} (-5.0^{\circ}\text{C})\\ &= -808 \,\text{kJ} \end{aligned}$

The heat removed from the water is $\boxed{- 808 \,\text{kJ}}$

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