Question

A freezer has a coefficient of performance of 6.30. It is advertised as using 486 kWh/yr. Note: One kilowatt-hour (kWh) is an amount of energy equal to running a 1-kW appliance for one hour.

a) On average, how much energy does it use in a single day?
I got 4.79 x 10^6 J

(b) On average, how much energy does it remove from the refrigerator in a single day?
I got 30177000 J

Now I'm stuck on this part:
What maximum mass of water at 17.1°C could the freezer freeze in a single day? (The latent heat of fusion of water is 3.33 x 10^5 J/kg, and its specific heat is 4186 J/kg · °C.) answer in kg

Answer 1

Compute the amount of heat that must be removed to freeze one kg of water that is initially 17.1 C. Assume the ice temperature stays at 0 C, once it is formed (although this is not necessarily the case)

That would be 3.35*10^5 J + 17.1*4186 = 4.046*10^5 J/kg

Finally, divide your computed heat removal by that heat per mass number.

Answer 2

Thank you!