# Latent heat and Phase Change

## Homework Statement

Trying to beat the heat of summer, a physics grad student went to the local toy store and purchased a child's plastic swimming pool. Upon returning home, he filled it with 156 liters of water at 26°C. Realizing that the water would probably not be cool enough, he threw ice cubes from his refrigerator, each of mass 30 g, into the pool. (The ice cubes were originally at 0°C.) He continued to add ice cubes until the temperature stabilized at 17°C. He 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.

How many ice cubes did he add to the pool to get the temperature to 17°C? (Consider the pool and ice cubes an isolated system.)

## Homework Equations

Q=mc(DeltaT)
Qwater/Qice=# of ice cubes

## The Attempt at a Solution

Mass water=1000*.156=156kg
Qwater=156000g(1.0 cal/g*degree Celsius)(17-26)=-1404000cal
QIce=30g(.5 cal/g*degree Celsius)(17-0)=255cal

now do i neglect the negative in Qwater? so then i would get 1404000cal/255cal=5505.88cubs but is not the correct answer

I don't understand what i have to do with the latent heat of H2O fusion(80cal/g)

Dick
Homework Helper
All of the cubes melted. Each gram of ice absorbed 80 cal/gm of heat as it melted and THEN the resulting WATER warmed up to 17 degrees. In fact, the heat capacity of ice has nothing to do with the problem.

All of the cubes melted. Each gram of ice absorbed 80 cal/gm of heat as it melted and THEN the resulting WATER warmed up to 17 degrees. In fact, the heat capacity of ice has nothing to do with the problem.

I still have no clue wtf is going on... now would i use the equation Q=mL for the ice? that would give me 2400cal... Then take the Qw=1404000/Qice=2400???? but that gives me 585, which is not the correct answer

Dick