A freezer is rated at 1100 watts. 90% of its power is used in the freezing process. How long will it take to make 3kg of ice at -10°C from water at 25°C?
3 kg of ice is equivalent to 3 kg of water because mass is conserved.
The freezer takes away heat by doing work.
So heat is taken away from 3 kg of water at 25 degrees to get 3 kg of water at 0 deg.
Then further heat is taken away from 3 kg of water to convert it to 3 kg of ice.
Finally heat is further taken away from 3 kg of ice at zero deg to get 3 kg of ice at -10 deg.
It is easier to understand if we reverse this problem, i.e. supply heat to 3 kg ice at -10 deg to convert it to 3 kg water at 25 deg using a heater.
The heat required for this process will be the same as the heat taken away in the previous case by the freezer.
The heat required will be
Q = 3 x Ci x [0 - (-10)] + (3 x Li ) + 3 x Cw x (25-0) where
Ci = Sp heat of ice, Li = Latent heat of fusion of ice, and
Cw = Sp heat of water. This result will be in calories. Convert to J.
The heater supplies (freezer takes away) Q J of energy at efficiency of 90% . Let E be the energy supplied to to the heater to get Q J of energy. Since the efficiency = 90%, we have Q/E = 0.9 or E = Q/0.9.
Now Power = energy / time So, Energy = Power x time So
E = 1100 x t. Solve for t.
Last edited by physicsquest; 06-01-2009 at 03:49 AM.
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