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

Answer:

$Q_r=334984.65\ J$

$E=111661.55\ J$

$TE=334984.65\ J$

$T_f\approx35^{\circ}C$

$COP=3$

Explanation:

Given:

• mass of water, $m_w=1\ kg$

• energy output to the energy input ratio of refrigerator, $E_{in}:E_{out}=3:1$

• mass of air in the kitchen, $m_a=40\ kg$

• specific heat of air, $c_a=0.2\ cal.g^{-1}.^{\circ}C^{-1}=837.2\ J.kg^{-1}.K^{-1}$

• specific heat of water, $c_w=4.186\ J.kg^{-1}.K^{-1}$

• latent heat of ice, $L=80\ cal.g^{-1}.^{\circ}C^{-1}=334880\ J.kg^{-1}.K^{-1}$

The amount of thermal energy to be removed from the 1 kg of water of room temperature to form ice:

$Q_r=m_w.c_w.\Delta T+m_w.L$

$Q_r=1\times 4.186\times (25-0)+1\times 334880$

$Q_r=334984.65\ J$

Electrical energy used to freeze the ice totally:

$E=\frac{Q_r}{3}$

$E=\frac{334984.65}{3}$

$E=111661.55\ J$ is the minimum electrical energy required.

Total energy including the waste energy rejected into the kitchen:

$TE=Q_r$ since we assume the system produces no extra energy.

$TE=334984.65\ J$

Rise in temperature of air in the kitchen:

$Q_r=m_a.c_a.\Delta T$

$334984.65=40\times 837.2\times (T_f-25)$

$T_f\approx35^{\circ}C$

Efficiency of the freezer is given by a parameter called the coefficient of performance:

$\rm COP=\frac{Desired\ effect}{energy\ consumed}$

as givne in the question:

$COP=\frac{3}{1}$

$COP=3$