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The density of water is 1000 kg/m3, the specific heat of water is 1.0 cal/g C, the specific heat of ice is 0.5 cal/g C, and the latent heat of fusion of water is 80 cal/g.

A) How many ice cubes did she add to the pool to get the temperature to 16 C? (Consider the pool and ice cubes an isolated system.)

So first I solved for heat lost by water...(200g)*(1 cal/g)*(25-16)= -1800 kj. I know that that has to equal the ice cubes heat lost.

to determine the ice cube temp change..i take a single ice cube and do the same procedures as before but adding the phase change... so its:(.5)(30)+(30)(1)(16-0)=495..

im just not sure if im going the right direction...if i am, i am stuck right here..