But what is :delta: T for the calorimeter?

Initial temperature of calorimeter was identical to that of water it contained.

But the heat capacity of the bomb calorimeter is 15.7 J/K without the water?

So if the final temperature of bomb calorimeter is 32.4 and the initial is 23.6, then :delta: T is 8.8 K,

So using the heat capacity (C) of 15.7 J/K, I can solve for q using q = C :delta: T

So q is 138.16 J

No, the :delta: T of water is also 8.8, so I can easily get the q.

But the q of the sample? Since q = m x Cs x :delta: T,

would....the :delta: T be 32.4 - 98.7 ?

And then I could just write the equation as

Cs x 307g x -66.3 K = -(q

_{water} + q

_{calorimeter})

And solve for Cs, right?

So Cs = [-(q

_{water} + q

_{calorimeter})]/(307g x -66.3 K)