Solving Heat Transfer of Ice Homework: Initial Copper Temp

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Bassa
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Homework Statement


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?[/B]

Homework Equations


Q=mcΔT
Q=mL (L in this case is the heat of fusion)
ΣQ=0[/B]

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?[/B]
 
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Chestermiller said:
You've made some arithmetic errors.

Chet
Thanks! Is the method correct, though?
 
Bassa said:

Homework Statement


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?[/B]

Homework Equations


Q=mcΔT
Q=mL (L in this case is the heat of fusion)
ΣQ=0[/B]

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?[/B]

You should check your arithmetic here throughout. For example, when I calculate 2 * 2100 * 20, I don't get 8400.
 
Your Latent heat of fusion of ice is incorrect. It should be 3.33 x 10^5 for it to have a result of 150 centigrade.