The refrigerator's temperature is -20 0C, we have put a 50 g of water of temperature 20.4 deg C letting some time to give 30 g of ice. Then, we are interested to calculate the energy dissipated to produce 30 g of ice
I am not sure what the question is? How much cooling is required to freeze 30g of water to ice? 79.72 cal/g *30g = 2391 cal is the simple answer. How much energy is required to perform that cooling? That depends on the refrigeration cycle efficiency. You didn't say anything about what temperature the water started at so there is probably some cooling additional cooling required to get to 0 C. I guess you also have 20g of water that you probably cooled to 0 C but didn't freeze yet. Its pretty easy to calculate how much cooling is required for that (about 1 cal/g deg C). Some of the ice will cool below 0 C eventually but probably not much if 20 g of water didn't freeze yet. What do you want to know? How long does it take? There is not really enough info given in the problem statement to calculate that.
Dear Fekadu,
Some information not given How much initial temp of water ?
In order for ice to go from an ice to a liquid state, energy must be added to cause the ice to go from a higher state to a lower ordered state.
The Heat dissipated to Water to Ice=M.L
Where M= mass of the ice
L= latent heat of Ice=79.72 cal/gm
i.e, 30 * 79.72=2391.6 cal or 2.3916 K. Cal
Sensible heat dissipate 0 to -20 degree C
M Cp. Dt = 30 * 1* 20=600 cal
Total Heat=2.3916 +0.6 K. cal =2392.2 K. Cal-------(1)
For energy calculation How much initial temp of water ?
20g of water that you probably cooled to 0 C but didn't freeze yet
The energy dissipated initial temp to zero degree of water
=M*Cp.dT
20 * 1 (t-0) -………(2)
Initial temp (t) unknown
30gm from t degree to zero degree c sensible heat
30*1 * (t-0) cal .......(3)
Total Heat = (1) + (2) +(3)
Regards,
Prem Baboo
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