- Download PDF
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
2 Answers | Add Yours
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.
plzzz discuss here
We’ve answered 319,500 questions. We can answer yours, too.Ask a question