Question

Question #26819

Answer 1

`119.0 J/(°C)`, or `C.`

Explanation:

In a constant-volume calorimeter, we relate the heat capacity of the calorimeter to the temperature change in said calorimeter with the equation:

`q_(soln)=-C_(cal)*DeltaT`

The mass of the substance in the calorimeter in this case is unnecessary; some professors put these data in problems to confuse students who don't really know their stuff. Oh, test curves...

Anyways,

`-250.0=-C_(cal)*2.1°C`
`C_(cal)=1.2*10^2 J/(°C)`

Note, the question says released, thus the sign for the heat (`q`) is negative, not positive. Moreover, in my actual calculation, `C.` is the answer; here I have truncated for clarity.