Question

An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 degrees Celsius and rejects heat to a room at a temperature of 20.6 degrees Celsius. Suppose that liquid water with a mass of 82.1 kg at 0.0 degrees Celsius is converted to ice at the same temperature. Take the heat of fusion for water to be Lf=3.34×105 J/kg.

A. How much heat |QH| is rejected to the room?
B. How much energy E must be supplied to the device?

Answer 1

A. Q = 2.74 x 10⁷ J = 27.4 MJ

B. E = 3.91 x 10⁸ J = 391 MJ

Step-by-step explanation:

A.

Heat rejected can be found as follows:

`Q = mL`

where,

Q = Heat rejected = ?

m = mass = 82.1 kg

L = Latent Heat of fusion = 334000 J/kg

Therefore,

`Q = (82.1\ kg)(334000\ J/kg)`

Q = 2.74 x 10⁷ J = 27.4 MJ

B.

First, we will calculate the efficiency of the Carnot Cycle as follows:

`Efficiency = 1-(T_1)/(T_2)`

where,

T₁ = Heat intake temperature = 0°C + 273 = 273 k

T₂ = Heat rejection temperature = 20.6°C + 273 = 293.6 k

Therefore,

`Efficiency = 1 - (273\ k)/(293.6\ k) \\\\Effciency = 0.07`

Therefore, the energy input required is:

`Efficiency = (Q)/(E)\\\\E = (Q)/(Efficiency) = (2.74\ x\ 10^7\ J)/(0.07)`

E = 3.91 x 10⁸ J = 391 MJ