# Chemisty-Heat capacity of calorimeter

1. Apr 10, 2010

### nwyatt

1. The problem statement, all variables and given/known data
6550 J of electrical heat is added to a constant pressure calorimeter containing 125 grams of water at 22.6 ◦C. If the final temperature of the calorimeter is 33.7 ◦C, the heat capacity of the calorimeter (including the water) is:

2. Relevant equations

I think it is either q=C*$$\Delta$$t or q= s*m*$$\Delta$$t

3. The attempt at a solution

I tried with the first equation and got 590 Joules/C and on the second I got 5800 Joules/C. Both of these answers were available as choices and I don't quite understand which one I would use and why. If anyone could help explain it would be great.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 11, 2010

### danago

If 6550J of energy is added, some of it will be used to heat the water, and some will be used to heat the calorimeter:

$$Q_{added} = Q_{water} + Q_{calorimeter}$$

Since you know the initial and final temperature of water, along with the mass, you should be able to calculate how much energy it used. Can you see what to do from here?

3. Apr 11, 2010

### Staff: Mentor

What is the definition of heat capacity? What it is that you are heating?

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