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Determine the ice cube temp change

  • Thread starter mattmannmf
  • Start date
  • #1
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Trying to beat the heat of the last summer, a physics student went to the local toy store and purchased a plastic child's swimming pool. Upon returning home, she filled it with 200 liters of water at 25oC. Realizing that the water would probably not be cool enough, she threw ice cubes from her refrigerator, each of mass 30g, into the pool. (The ice cubes were originally at 0oC.) She continued to add ice cubes, until the temperature stabilized to 16oC. She then got in the pool.

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..
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
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200 liters H2Oski = 200*1000g = 200,000 g.

Consider that you have 200,000 g of water.

You have the heat given up by the water = the heat required to melt ice and raise its temp to stasis

200,000g*1*9 = X*(80 + 16)*30

where X is the # of cubes.
 

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