Chemisty-Heat capacity of calorimeter

[nwyatt]

Homework Statement

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:

Homework Equations

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

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.

 

[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?

 

[Borek]

the heat capacity of the calorimeter (including the water)

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

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