# Heat Transfer of ice

## 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
what did I do wrong?[/B]

Last edited:

Related Introductory Physics Homework Help News on Phys.org
Chestermiller
Mentor

Chet

Chet
Thanks! Is the method correct, though?

Chestermiller
Mentor
Thanks! Is the method correct, though?
Yes

SteamKing
Staff Emeritus
Homework Helper

## 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