Refrigerator power rating and heat absorption

AI Thread Summary
A refrigerator rated at 370 W can absorb heat from the food compartment, but the calculation requires understanding the Coefficient of Performance (COP) and the relationship between heat absorbed (Qc) and work done (W). The COP equation, COP = Qc/W, is crucial for determining the maximum heat absorption, where W represents the power rating in Joules per second. The user is confused about how to apply the equations, particularly regarding the temperatures of the hot and cold reservoirs, and whether all the power is used for work. Clarification on these concepts is necessary for solving the problem effectively. Understanding these principles will enable accurate calculations for heat absorption in refrigerators.
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Homework Statement



A refrigerator is rated at 370 W. What is the maximum amount of heat it can absorb from the food compartment in 1 minute if the food compartment temperature is 0 deg C and it releases heat into a room at 20 degrees C.

Homework Equations



COP=Qc/W

W=Qh-Qc

The Attempt at a Solution



Now I have several questions about this one because my book does not expain at all how to do a problem like this. First of all I know that 370 W means 370 J/s but does that mean that all of that 370 Joules of energy is turned into work in trying to move heat from the cold reservoir to the hot reservoir? Also the only equations that I know are the ones listed above but I'm not even sure tht the second one is relevant because it is for heat engines not refrigerators. The book doesn't give a second equation for refrigerators. So how do I relate the temperatures of the different reservoirs to the Qh and Qc because this was not discussed either?

Pleeeeeeeeease help!
 
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COP=\frac{T_H}{T_H - T_C}

TH is the temp of the ambient surrounding
TC is the temp of the fridge.
 
Oh yes thank you
 
is W equal to the power rating?
 
I'd believe so,yes.
 
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