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Question

A freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80 kg of water at 25.0C25.0^{\circ} \mathrm{C} to 1.80 kg of ice at 5.0C-5.0^{\circ} C in one hour. (a) What amount of heat must be removed from the water at 25.0C25.0^{\circ} C to convert it to ice at 5.0C-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

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Answered 11 months ago
Answered 11 months ago
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Given

We are given a coefficient of performance KK = 2.40 for a freezer. A mass of water mm =1.80 kg at temperature Tw=25.0CT_{w} = 25.0^{\circ} \text{C} convert with the same mass to ice and reach the temperature Tice=5.0CT_{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 25.0C25.0^{\circ}\text{C} to 0.0C0.0 ^{\circ} \text{C} so the heat in this process will be given by

Q=mcwΔT=mcw(0.0C25.0C)(1)\begin{aligned} Q &= mc_{w}\Delta T = mc_{w} (0.0^{\circ}\text{C}- 25.0^{\circ} \text{C}) \tag{1} \end{aligned}

Where cwc_{w} is the specific heat capacity of water and equals 4190 J/kgCo\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=mLf(2)Q = -mL_{f} \tag{2}

Where LfL_{f} is the heat fusion of water and equals 3.34 ×105\times 10^{5} J/kg. (See Table 17.4)
The third process, the ice temperature changes from 0.0C0.0^{\circ} \text{C} to 5.0C-5.0^{\circ} \text{C} and the heat in this process will be given by

Q=mciceΔT=mcice(5.0C0.0C)(3)\begin{aligned} Q &= mc_{ice}\Delta T = mc_{ice} (-5.0^{\circ}\text{C}- 0.0^{\circ} \text{C}) \tag{3} \end{aligned}

Where cicec_{ice} is the specific heat capacity of the ice and equals 2100 J/kgCo\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)

Q=mcw(25.0C)mLf+mcice(5.0C)=1.80kg×4190 J/kgCo(25.0C)1.80kg×3.34×105J/kg+1.80kg×2100 J/kgCo(5.0C)=808kJ\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 808kJ\boxed{- 808 \,\text{kJ}}

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