# Homework Help: Specific heat of a piece of metal dropped in water

1. Nov 16, 2007

### cyclonefb3

1. The problem statement, all variables and given/known data

A metal container, which has a mass of 8.0 kg contains 14.3 kg of water. A 2.9-kg piece of the same metal, initially at a temperature of 190.0°C, is dropped into the water. The container and the water initially have a temperature of 15.1°C and the final temperature of the entire system is 17.1°C. Calculate the specific heat of the metal.

2. Relevant equations

Q = c* m *change in T

3. The attempt at a solution

The heat of the container + the heat of the water = the heat of the piece of metal

using algebra I got the equation:

cwater*Mwater*change in Twater / [Mpiece*change in Tpiece - Mcontainer*change in Tcontainer]

2. Nov 16, 2007

### mgb_phys

Nearly right, The change in T for the container and the water is the same.

So the equation is
Cwater * Mwater * dT water + Cmetal * MContainer * dTwater = Cmetal * Mmetal *dTmetal

The only unknown is Cmetal.

3. Nov 16, 2007

### Shooting Star

Hi cyclonefb3,

Let t be the final temp. Then,

(14.3*Cw+8*Cm)(17.1-15.1) = (2.9*Cm)(190-17.1).

If Cwater is known, you can find Cm.

4. Nov 16, 2007

### ttk3

I think I have the dTpiece wrong this is what i'm entering:

41.86(14.3)(17.1-15.1) / [2.9*(190-15.1)] - [8*(17.1-15.1)]

I'm not sure where to go from here...

5. Nov 16, 2007

### mgb_phys

Cwater * Mwater * dT water + Cmetal * MContainer * dTwater = Cmetal * Mmetal *dTmetal
4200 * 14.3 (17.1-15.1) + C * 8 (17.1-15.1) = C * 2.9 (190-17.1)
120120 = C (2.9*172.9 - 8*2)

Hint C for metals is generally a few hundred J/kg/k

Last edited: Nov 16, 2007
6. Nov 16, 2007

### ttk3

thanks, it was just he program being really picky with digits.