Carnot refrigerator cost problem

[RockJockey]

Homework Statement

Three kilograms of liquid water at 0° C is put into the freezer compartment of a Carnot refrigerator. The temperature of the compartment is -13.5° C, and the temperature of the kitchen is 24.5° C. If the cost of electrical energy is ten cents per kilowatt · hour, how much does it cost to make two kilograms of ice at 0° C?

Homework Equations

COP= Qc/W , Qc/Qh=TL/Th , Q=mL L=3.35 x 10^5 j/kg , Qh= W+ Qc, 3,600,000 = 1 kilowatt hour

The Attempt at a Solution

I used the formula Q=mL to solve for Qc, because this is the heat that needs to be moved to the hot reservoir, got an answer of Qc=1005000J. I then used Qc/Qh = TL/Th to solve for Qh came out to be 1149885.584j. I then used the Qh=W+Qc to solve for W, which equaled 144885.584j. I then solved for the COP = 6.937. I did convert temp values to K. I'm lost after this point I need to know how to come up with the cost.

 

[collinsmark]

Homework Statement

Three kilograms of liquid water at 0° C is put into the freezer compartment of a Carnot refrigerator. The temperature of the compartment is -13.5° C, and the temperature of the kitchen is 24.5° C. If the cost of electrical energy is ten cents per kilowatt · hour, how much does it cost to make two kilograms of ice at 0° C?

Homework Equations

COP= Qc/W , Qc/Qh=TL/Th , Q=mL L=3.35 x 10^5 j/kg , Qh= W+ Qc, 3,600,000 = 1 kilowatt hour

The Attempt at a Solution

I used the formula Q=mL to solve for Qc, because this is the heat that needs to be moved to the hot reservoir, got an answer of Qc=1005000J.

The problem statement says that you only need to make 2 kg of ice. Even though you put 3 kg of water in the freezer, you only need to make 2 kg of ice (the other 1 kg can remain as water). :tongue2: So you'll have to redo your Q_c calculation.

I then used Qc/Qh = TL/Th to solve for Qh came out to be 1149885.584j. I then used the Qh=W+Qc to solve for W, which equaled 144885.584j. I then solved for the COP = 6.937.

What I would have done is first solved for the COP based on T_hot and T_cold. Once you have the COP, you can easily solve for the work W using your COP= Q_c/W formula. My way is a little easier, but your method works just fine too. But either way, you'll have to redo the calculations using only 2 kg of ice (instead of 3) this time.

I'm lost after this point I need to know how to come up with the cost.

Once you find the work in Joules, you have to convert that to kilowatt · hours.

Joules are a measure of energy. kilowatt · hours are also a measure of energy (just different units).

Here is a hint. 1 Watt = 1 Joule/second.