A freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80Kg of water at 25.0◦C to 1.80kg of ice at -5.0◦C in hour. A freezer has a coefficient of performance of 2.40....

A freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80Kg of water at 25.0◦C to 1.80kg of ice at -5.0◦C in hour.

A freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80Kg of water at 25.0◦C to 1.80kg of ice at -5.0◦C in hour. 

a)      What amount of heat must be removed from the water at 25.0 ◦C to convert it to ice at -5.0◦C?

b)       How much electrical energy is consumed by the freezer during this hour?

c)       How much wasted heat is delivered to the room in which the freezer sits?

Note that:     for water Cw = 4190J/Kg.K, Lf = 3.34 × 105J/Kg.K & Cice = 2010J/Kg.K

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Your question belongs in the Science Q+A part of eNotes but I'll discuss it anyway.

COP. Your freezer takes 100 watts to make 240 watts . Incidentally breaking the Second Law of Thermodynamics - but not really.

You need to know two important equations.

When something changes its temperature but doesn't change phase the equation is:

Q = mc(T2 - T1)

Sorry I can't use super/subscript here.

During ( and only during) melting or boiling:

Q = mL.

Pleased to solve your simple heat calculation problem in another forum.

Steve K.

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