A Carnot freezer that runs on electricity removes heat from the freezer compartm
ID: 1505906 • Letter: A
Question
A Carnot freezer that runs on electricity removes heat from the freezer compartment, which is at -10.0 C and expels it into the room at 29.0 C. You put an ice-cube tray containing 385 g of water at 16.0 C into the freezer.
What is the coeffiecient of performance of this freezer?
How much energy is needed to freeze the water?
How much electrical energy must be supplied to the freezer to freeze the water?
How much heat does the freezer expel in to the room while freezing the ice?
Please explain how to find Cw, Cice, and etc. I am very confused on this topic.
Explanation / Answer
specific heat of water is 4.186 kJ/kgC
specific heat of ice is 2.06 kJ/kgC
heat of fusion of ice is 334 kJ/kg
energy or heat needed = mass * specific heat * change in temperature
energy needed to cool water to 0 degree E1 = 4.186 * 0.385 * (16 - 0)
E1 = 25.78576 J
energy needed to make ice E2 = 334 * 0.385
E2 = 128.59 J
energy needed to cool ice to -10 degree C E3 = 2.06 * 0.385 * (0 - (-10))
E3 = 7.931 J
energy needed to freeze water E = E1 + E2 + E3
energy needed to freeze water E = 25.78576 + 128.59 + 7.931
energy needed to freeze water E = 162.30676 J
Efficiency of a carnot cycle heat engine = 1 – (Tc / Th)
Where Tc is absolute temperature of cold reservoir
Where Th is absolute temperature of not reservoir
= 1 – (263.15 / 302.15)
= 0.129
electrical energy must be supplied to the freezer to freeze the water = E /
electrical energy must be supplied to the freezer to freeze the water = 162.30676 / 0.129
electrical energy must be supplied to the freezer to freeze the water = 1258.19 J
The Coefficient of Performance (COP) of the heat engine = 1 /
The Coefficient of Performance (COP) of the heat engine = 1 / 0.129
The Coefficient of Performance (COP) of the heat engine = 7.75
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