Which Uses Less Energy: Refrigerator or Freezer for Cooling Water?

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Homework Help Overview

The discussion revolves around comparing the energy efficiency of a refrigerator and a freezer when cooling a specific volume of water from room temperature to a lower temperature. The original poster presents a scenario involving a refrigerator set at +2 °C and a freezer at -18 °C, both operating in a room at +20 °C, and seeks to determine the relative electric energy consumption for this cooling process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the coefficient of performance (C.O.P) for both the refrigerator and freezer, as well as the work required to cool the water. There is confusion regarding the interpretation of energy versus power and how to apply the equations correctly to find the work done in each case.

Discussion Status

Some participants have provided guidance on applying the equations to calculate work based on the C.O.P values. There is acknowledgment of the efficiency differences between the refrigerator and freezer, but confusion remains regarding the implications of these calculations and the units of measurement required for the final answer.

Contextual Notes

Participants note the importance of understanding the energy withdrawal from the water and clarify that the problem specifically asks for energy in kilojoules rather than power in watts. There is an emphasis on the ideal operation of both systems, which may influence the assumptions made during calculations.

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Homework Statement


Compare a refrigerator set at +2 oC and a freezer set at -18 oC, both of which are operating in the room with temperature of +20 oC. What is relative amount of electric energy that will need to be used in both cases to cool a cup of water (250 ml) from room temperature to +5 oC? Assume ideal operation of both systems.


Homework Equations



C.O.P = Tcold/(Thot-Tcold)
C.O.P = Qc/w
Q = mCΔT

The Attempt at a Solution



C.O.P (refrigerator) = 275K/(293K-275K) = 15.278
C.O.P (freezer) = 255K/(293K-255K) = 6.711

Q = 250g(4.1855 J/g*K)(278-293K) = -15.696kJ
 
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I would be tempted to say you are almost there and so all you need to do is to apply you second equation (in each case ) using the cop for thereby calculating w since you have worked out the Q cold value.
 
If I just applied the 3rd equation I would be calculating work in kJ and the amount of work for a refrigerator to cool the water would be lower than for a freezer, which confuses me. Also the problem asks for electric energy in J/s or Watts.

(refrigerator) w = Qc/COP = -15.696 kJ/15.278 = -1.0274 kJ
(freezer) w = Qc/COP = -15.696 kJ/6.711 = -2.3388 kJ
 
I think the purpose of giving you the info abt the cup of water us to allow you to calculate the energy to be withdraw. You correctly calculated thus as 15.7 kJ. The fridge is more efficient with better cop, so you need just 15.7/15.278 kJ while the freezer needs 15.7/6.711 kJ.

The relationships look ok -- for pumps and refris lower temp differential better, for heat engines the opposite. Finally the qs as you have stated it is asking for energy not power so KJ is correct and not watts.
That's my take anyway.
 

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