Calculate Work and Entropy Change for an Ice Maker

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An ice maker inputs liquid water at 25 degrees C and outputs ice at -5 degrees C. Assume there is 1 kg of water and the volume does not change.

Cp liquid 4.18 kJ/kg-K
Cp solid 2.11 kJ/kg-K
∆H fusion 334 kJ/kg

I need to calculate:

  1. Entropy change of the water, in kJ/K
  2. Minimum electrical work required, in kJ
  3. Allowing the ice to thaw and return to room temperature, what is the universe's change in entropy for the cycle, in kJ/K?
  4. What about if the ice maker has an efficiency of 20%?

I've already calculated the water's entropy change as -1.628 kJ/K. Where would I go from here, for the second question? Would work be the difference between Qhot and Qcold, where Q is the change in entropy times the temperature in K?

I've calculated the coefficient of performance (COP) to be 8.9383333 (or is it 0.100620493?), but I'm unsure of how to use that to find the work.

Any help will be greatly appreciated.
 
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The entropy change of the water is -1.628 kg/K. This is the negative of the entropy transferred from the water to the refrigerator cycle on the cold side. If the refrigerator operates ideally, this same amount of entropy is transferred from the refrigerator cycle to the room (at room temperature) on the hot side. If the room is at 25 C, what is the amount of heat removed from the refrigerator on the hot side? How much heat is removed from the water on the cold side?
 
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