# Heat Transfer of ice

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1. Nov 27, 2014

### Bassa

1. The problem statement, all variables and given/known data
A 6.00-kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 1.2 kg of ice and 0.80 kg of liquid water. What was the initial temperature od the piece of copper?

2. Relevant equations
Q=mcΔT
Q=mL (L in this case is the heat of fusion)
ΣQ=0

3. The attempt at a solution
Qcopper= 6(390)(Tf-T)
Qice=2(2100)(0-(-20))=8400J (heat required to raise the temperature of the ice to 0C)
Qwater=.8(334x10^3)=26400J (heat required to melt .8 kg of ice)

applying ΣQ=0:
6(390)(Tf-T) + 8400+26400=0

Tf= would have to be 0c because we still have some ice left.

solve for T:
T
=118.974C
The correct answer is 150C
what did I do wrong?

Last edited: Nov 27, 2014
2. Nov 27, 2014

### Staff: Mentor

You've made some arithmetic errors.

Chet

3. Nov 27, 2014

### Bassa

Thanks! Is the method correct, though?

4. Nov 27, 2014

### Staff: Mentor

Yes

5. Nov 27, 2014

### SteamKing

Staff Emeritus
You should check your arithmetic here throughout. For example, when I calculate 2 * 2100 * 20, I don't get 8400.

6. Nov 27, 2014

### Bassa

Thank you very much for your help!

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