Question

 We want to determine how much the room temperature increases when a kg of ice freezes.

Suppose you have a freezer that needs 1 J of energy for every 3 J of heat it removes.

How much thermal energy must be removed from 1 kg of water at room temperature? How much electrical energy is used to freeze the ice?

What is the total energy, including waste heat, that is dumped into the kitchen? If the kitchen contains 40 kg of air, how much will the temperature rise? What is the efficiency of the freezer?

(The latent heat of water for freezing is 80 cal/gm°C, and the specific heat of air is about 0.2 cal/gm°C.  We know you could look it up; we thought we'd save you the time.)

Answer 1

1)

Mass of water =1kg

Temperature of water \(\mathrm{T}=20^{\circ} \mathrm{C}\) (room temperature)

Amount of Heat required to be removed to convert water at \(20^{\circ} \mathrm{C}\) into ice

= Heat required to bring water to \(0^{\circ} \mathrm{C}+\) Heat required to convert it into ice at \(0^{\circ} \mathrm{C}\)

\(=\left[\mathrm{m}_{w}\left(4.168 \frac{\mathrm{J}}{\mathrm{gm}^{\circ} \mathrm{C}}\right)(\Delta \mathrm{T})\right]+\left[\mathrm{m}_{w}\left(334 \frac{\mathrm{J}}{\mathrm{gm}}\right)\right]\)

\(=\left[1000 \mathrm{gm}\left(4.168 \frac{\mathrm{J}}{\mathrm{gm}^{\circ} \mathrm{C}}\right)\left(20^{\circ} \mathrm{C}\right)\right]+\left[1000 \mathrm{gm}\left(334 \frac{\mathrm{J}}{\mathrm{gm}}\right)\right]\)

$$ =83360 \mathrm{~J}+334000 \mathrm{~J} $$

\(=417360 \mathrm{~J}\)

\(=41736 \mathrm{~kJ}\)

Therefore, the amount of heat required to be removed by the refrigerator is \(Q=417.36 \mathrm{~kJ}\)

2)

As the refrigerator needs \(1 \mathrm{~J}\) of electrical energy to remove \(3 \mathrm{~J}\) of heat

Electrical energy required to remove \(\mathrm{Q}=417360 \mathrm{~J}\) heat is

Electrical energy required \(=\frac{Q}{3}\)

$$ \begin{array}{l} =\frac{417360 \mathrm{~J}}{3} \\ =139120 \mathrm{~J} \\ =139.12 \mathrm{~kJ} \end{array} $$

Therefore, total electrical energy required is \(139.12 \mathrm{~kJ}\)

3)

Total Heat that is Dumped into the kitchen is due to the heat taken out from water and the heat input to the refrigerator.

Heat \(_{\text {Room }}=\) Heat \(_{\text {Water } \rightarrow \text { ice }}+\) Heat \(_{\text {electical }}\)

Heat \(_{\text {Room }}=417360 \mathrm{~J}+139120 \mathrm{~J}\)

Heat \(_{\text {Room }}=556480 \mathrm{~J}\)

Heat \(_{\text {Room }}=556.48 \mathrm{~kJ}\)

Therefore, total heat energy dumped to the kitchen is \(556.48 \mathrm{~kJ}\)

4)

The total heat radiated to the room will increase the temperature of air inside the room

The total heat radiated is \(556480 \mathrm{~J}\)

The total heat radiated in calories \(=133001.912\)

The initial temperature of air is \(20^{\circ} \mathrm{C}\)

The mass of air inside room is \(40 \mathrm{~kg}\)

Specific heat capacity of air \(0.2 \frac{\mathrm{cal}}{\mathrm{gm}^{\circ} \mathrm{C}}\)

Therefore, the temperature rise is

\(\mathrm{Q}=\mathrm{m}_{\mathrm{a}} \mathrm{C}_{\mathrm{V}}\left(\mathrm{T}_{2}-20^{\circ} \mathrm{C}\right)\)

\(133001.912=(40,000 \mathrm{gms})\left(0.2 \frac{\mathrm{Cal}}{\mathrm{gm}^{0} \mathrm{C}}\right)\left(\mathrm{T}_{2}-20^{\circ} \mathrm{C}\right)\)

\(16.62^{\circ} \mathrm{C}=\left(\mathrm{T}_{2}-20^{\circ} \mathrm{C}\right)\)

\(\mathrm{T}_{2}=36.62^{\circ} \mathrm{C}\)

Therefore, the temperature will rise by \(16.62^{\circ} \mathrm{C}\), and the final temperature of the air will be \(\mathbf{T}_{2}=36.62^{\circ} \mathrm{C}\)

5)

The efficiency of refrigerator is

\(\mathrm{COP}_{\mathrm{R}}=\frac{\text { Output }}{\text { Input Energy }}\)

\(\mathrm{COP}_{\mathrm{R}}=\frac{417.36 \mathrm{~kJ}}{139.12 \mathrm{~kJ}}\)

\(\mathrm{COP}_{R}=3\)