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 = -(qwater
And solve for Cs, right?
So Cs = [-(qwater
)]/(307g x -66.3 K)