# Homework Help: Calorimeter heat capacity

1. Jan 13, 2008

### Hoejer

1. The problem statement, all variables and given/known data
A copper calorimeter with the heatcapacity of 75 J/K has 300 g of petrolium in it. The starttemp. is 17.8 C. A kobberweight of 100 g. is 100Celcius. It is being put in the petroleum, where the temperature ends up being 22 C. The specifik heatcapacity of coppe ris 387J(kg/K)
What is petroleums specifik heat capacity?

2. Relevant equations

C_pet*m_pet(t2-ts)+C_cop*m_cop(t2-t1)+C_cal*m_cal(t2-t1)=0

3. The attempt at a solution
I think I have overlooked something - so I just need hints on the way.

I don't have the mass of the calorimeter, but I now that it's heatcapacity is 75 J/K and I know that it's made of coper and the specifik heat capacity of that is 387 J(kg*K)

If that is the way to go, then I have all the variables except the specifik heat capacity of petroleum, and then I just isolate C-Pet and it's piece of cake!

2. Jan 13, 2008

### Hoejer

the the mass of the calorimeter must be 75/387 = 0.193 kg = 193 g

Which then means C_pet*300(22-100)+387*100(22-17.8)+387*193(22-17.8)=0

C_pet = 20.4 J(kg/K)

3. Jan 13, 2008

### Hoejer

Which isn't true as looking in table it is around 1900 J/(kg*K)

4. Jan 14, 2008

No help?