# Heat loss in a system

1. Jul 2, 2012

### sgstudent

1. The problem statement, all variables and given/known data
A plastic ice tray has a section filled with water. When placed in a coler, the water freezes into ice cubes. The mass of water is 320g and the initial temperature of water is 0 degrees Celsius. The specific latent heat of fusion is 330J/g. Pipes containing fluids are used to take away heat from the cooler.
A)calculate the amount of heat required to freeze the water.
B)state why the heat capacity of the plastic tray does not affect the answer to A).

2. Relevant equations
Q=ml

3. The attempt at a solution
A)320x330=105600J

B) is the part that I'm more concerned with. The answer was that the temperature of the plastic tray and the water remains the same. Hence no heat is absorbed by the plastic tray and the heat capacity does not affect the answer in A).

I don't understand this. Could someone explain this to me? Thanks for the help!

2. Jul 2, 2012

### Infinitum

Hi!

What about this did you not understand? Heat capacity comes into picture only when the temperature changes. But when water is cooled to ice(phase change) there is no change of temperature.

3. Jul 2, 2012

### TaxOnFear

When a phase change occurs, heat energy is used to break or make bonds in the structure (absorb or release energy respectively). The energy is not effecting the internal energy of the substance therefore the temperature remains the same throughout the change.

4. Jul 2, 2012

### sgstudent

But if the water is at 10 degree Celsius then how will the energy given out to turn into 0 degrees water? I don't get how the specific heat capacity will affect the energy required to cool down then. In my first question, heat is given out from the 0 degrees water but heat is still given out so I'm confused how it should be affected even when temperature does change.

Btw for taxonfear, internal energy changes as the potential energy increases.

Thanks for the help!

5. Jul 2, 2012

### Infinitum

Not sure what you mean

But if the water and tray are both at 10 degrees, then that temperature would have had to be brought to zero first, before freezing occurs. If the tray was at 10 degrees, and water at 0 degrees, then the tray would cool down to 0, after which, the water starts freezing.

6. Jul 2, 2012

### sgstudent

I suddenly understood it. If the plastic tray was at 10 degrees and water at 0, then heat will enter the water hence more heat will have to be taken out. But if the plastic tray was at -10 degrees then less heat will be taken out? But won't the energy taken out from the water when the plastic tray is -10 degrees still be the same? Since energy is given out to both the tray and the fluid, but the net given out is still the same whereas in the 10 degrees tray, more heat must be taken out since some heat also enters the water?

7. Jul 2, 2012

### Infinitum

No. The plastic and water try to establish thermal equilibrium between themselves too, so the water being at a lower temperature transfers heat to the tray to raise its temperature, till they both finally have the same temperature. By then, the water has already lost some heat, so the amount of external heat needed to freeze it will obviously be less than the case where both are at 0 degrees.

8. Jul 2, 2012

### sgstudent

Oh, so we measure by using how much energy is taken away by the external?