# Copper calorimeter problem

1. Sep 10, 2009

### amiv4

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

A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure.

2. Relevant equations

If 0.750 kg of lead at a temperature of 255 C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

3. The attempt at a solution

mc$$\Delta$$tcopper+mc$$\Delta$$tice+water+mL(ice)=mc$$\Delta$$tlead

.1(390)(Tf-273.15)+.178(4190)(Tf-273.15)+(.018)334x103=.750(130)(Tf-528.15)

2. Sep 10, 2009

### kuruman

Re: Calorimetry

The temperature change of the lead is negative while the temperature changes of the copper and water are positive. Since you put the heat term from the lead on the right side of the equation, you need to write the temperature difference as 528.15 - Tf, otherwise you have a whole lot of positive numbers on the left equal to a negative number on the right. You can also leave it as is, but then it must be on the left side, so that the sum of all heat losses/gains is equal to zero.

3. Sep 10, 2009

### amiv4

Re: Calorimetry

I moved everything and game up with 308.14 K and then changed that to 34.99 C but that wasn't the right answer

4. Sep 10, 2009

### kuruman

Re: Calorimetry

I get 21.4 oC. Is that the answer? If it is, then recheck your work. Also, since all you have is temperature differences ΔT, you can do your calculations in degrees Celsius and forget about subtracting the 273.15. A change in temperature is the same in Celsius as it is in Kelvin. This will simplify your calculation because you will have Tf instead of (Tf - 273.15) in your expression.

5. Sep 10, 2009

### amiv4

Re: Calorimetry

Thank you I must have put stuff in my calculator wrong