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## 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 h

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?

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Physics
5 months
2021-08-04T06:51:08+00:00
2021-08-04T06:51:08+00:00 1 Answers
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## Answers ( )

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:

where,

Q = Heat rejected = ?

m = mass = 82.1 kg

L = Latent Heat of fusion = 334000 J/kg

Therefore,

Q = 2.74 x 10⁷ J = 27.4 MJB.First, we will calculate the efficiency of the Carnot Cycle as follows:

where,

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

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

Therefore,

Therefore, the energy input required is:

E = 3.91 x 10⁸ J = 391 MJ