Determine the ice cube temp change

[mattmannmf]

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 I am going the right direction...if i am, i am stuck right here..

 

[LowlyPion]

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.

 

[nlightner]

I would approach this problem by first calculating the total heat lost by the water in the pool. This can be done by using the equation Q = mcdeltaT, where Q is the heat lost, m is the mass of water, c is the specific heat of water, and deltaT is the change in temperature. Using the given values, we can calculate the heat lost by the water to be 1800 kJ.

Next, I would calculate the heat lost by each individual ice cube. This can be done by using the same equation, but with the specific heat of ice and the change in temperature from 0oC to 16oC. The heat lost by each ice cube can be found to be 495 J.

To determine the number of ice cubes added, we can divide the total heat lost by the heat lost by each ice cube. This gives us 1800 kJ / 495 J = 3.64 ice cubes. Since we cannot have a fraction of an ice cube, we can round up to 4 ice cubes.

Therefore, the student added 4 ice cubes to the pool to lower the temperature to 16oC. This calculation assumes that the pool and ice cubes are an isolated system and there are no other factors affecting the temperature change.