COP refrigerator/freezer problem

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In summary, the refrigerator will use more electric energy to cool a cup of water from room temperature to 5 oC than the freezer.
  • #1
Pauli_Pocket
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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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5


W = Q/C.O.P
W (refrigerator) = -15.696kJ/15.278 = -1.026kJ
W (freezer) = -15.696kJ/6.711 = -2.336kJ

The relative amount of electric energy needed for the refrigerator is less than the freezer, as the C.O.P for the refrigerator is higher. This means that for the same amount of energy input, the refrigerator is able to produce more cooling effect than the freezer. Therefore, the refrigerator will use less electric energy than the freezer to cool the cup of water from room temperature to +5 oC.
 

1. What is a COP refrigerator/freezer problem?

A COP (Coefficient of Performance) refrigerator/freezer problem refers to an issue with the efficiency of the refrigeration or freezing process in a refrigerator or freezer. This can result in inadequate cooling or freezing, leading to spoiled food or increased energy consumption.

2. What are some common causes of a COP refrigerator/freezer problem?

Some common causes of a COP refrigerator/freezer problem include a malfunctioning compressor, faulty or dirty condenser coils, a low refrigerant level, or a defective thermostat. These issues can prevent the refrigerator/freezer from maintaining the proper temperature and can also put strain on the unit's energy efficiency.

3. How can I troubleshoot a COP refrigerator/freezer problem?

To troubleshoot a COP refrigerator/freezer problem, you can start by checking the temperature settings and making sure they are set correctly. Next, clean the condenser coils to ensure proper air flow, and check the door seals for any cracks or damage. If these steps do not solve the issue, it may be necessary to call a professional for further diagnosis and repair.

4. Can a COP refrigerator/freezer problem be fixed?

In most cases, a COP refrigerator/freezer problem can be fixed. However, the extent of the repair will depend on the specific issue causing the problem. In some cases, a simple adjustment or cleaning may solve the problem, while more serious issues may require replacement of parts or even the entire unit.

5. How can I prevent a COP refrigerator/freezer problem from occurring?

To prevent a COP refrigerator/freezer problem, it is important to regularly clean and maintain the unit. This includes cleaning the condenser coils, checking the door seals, and keeping the temperature settings at the recommended levels. It is also important to address any issues or malfunctions as soon as they arise to prevent further damage and maintain the efficiency of the unit.

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