# Homework Help: Calorimetry, calculating for mass

1. Nov 8, 2007

### yoshi-chan7

Hi, here is the problem that I'm having trouble with.

It is desired to cool iron parts from 600°F to 200°F by dropping them into water that is initially at 55°F. Assuming that all the heat from the iron is transferred to the water and that none of the water evaporates, how many kilograms of water are needed per kilogram of iron?

I've used this equation:

mass of iron parts * specific heat of iron * change in temp. of iron parts = mass of water * specific heat of water * change in temp. of water

I first calculated the change in temperature of the iron parts and converted it to degree Celsius by 200F - 600F = 400F and then 400F*(5/9), which yields -222.

I plugged in the values as thus:

m*448 J/(kgC)* -222C = m*4186 J/(kgC)* 222C.

I've got a sneaking suspicion that it's because of the right side of my equation dealing with water. I think that somehow my scale is wrong, the 222C. But if all the heat released by the iron is transferred to the water with no loss, then the scale would be correct, wouldn't it?

2. Nov 8, 2007

### rl.bhat

Heat lost by the iron part = heat gained by the water.
1*448 J/(kgC)* 222C = m*4186 J/(kgC)* 80.6C.

3. Nov 8, 2007

### yoshi-chan7

I don't understand how you got the 80.6C. Could you explain that, please?

4. Nov 8, 2007

### yoshi-chan7

And also, why did the mass of the iron part become merely "1"? Wouldn't the mass be unknown?

5. Nov 9, 2007

### rl.bhat

how many kilograms of water are needed per kilogram of iron?
Initial temperature of water 55F and final teperature of water is 200F
Convert this difference into C

6. Nov 9, 2007

### yoshi-chan7

Thank you very much! I simply forgot to apply the formula for conversion of temperature units for the change in temperature. I understand perfectly now. Thank you~!