Hard Physics Question(heat)??

Question

An aluminium can of mass 100g contains 200g of water>Both initially at 15 degrees celsius, are placed in a refrigerator at -5.0 degrees celsius.Calculate the quantity of heat that has to be removed from the water and the can for their temperature to fall to -5.0?

Answer 1

for water(from 15 celsius to 0 celsius)

initial temp. = 15 celsius

final temp. = 0 celsius

specific heat capacity=1cal/g*celsius

heat=200g * 1cal/g*celsius * 15 celsius

=3000cal=3 k cal

from water to ice

heat=mass*latent heat of fusion

=200g * 80 cal/g

=16000cal=16 k cal

from 0 celsius to -5 celsius

heat=200g * 1cal/g*celsius * 5 celsius

=1000cal=1 k cal

total heat LOST=(3+16+1) k cal

=20 k cal

For aluminum you require its specific heat capacity and multiply it with the mass and change in temp. you will get your ans.

Answer 2

you will need some constants for that calculation, The water has to change to zero C, then freeze, then continue to drop, the aluminum could be done in one step.