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

Answer:

A. Q = 2.74 x 10⁷ J = 27.4 MJ

B. E = 3.91 x 10⁸ J = 391 MJ

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:

$Efficiency = 1-\frac{T_1}{T_2}$

where,

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

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

Therefore,

$Efficiency = 1 - \frac{273\ k}{293.6\ k} \\\\Effciency = 0.07$

Therefore, the energy input required is:

$Efficiency = \frac{Q}{E}\\\\E = \frac{Q}{Efficiency} = \frac{2.74\ x\ 10^7\ J}{0.07}$

E = 3.91 x 10⁸ J = 391 MJ