[Ben Cohen]

Hey guys, I've been having trouble with this problem:

A 9.40 g sample of KBr is dissolved in 105g of H2O at 23.6 °C in a coffee cup. Find the final temperature of this system. Assume that no heat is transferred to the cup or the surroundings.

Molar mass KBr 119 g/mol
ΔHsoln KBr 19.9 kJ/mol
Cp solution = 4.184J/g.C

So I figured this was:
q(KBr) = - q (H2O)
(9.4g)(1mol/119g)(19900J/mol) = -(105g)(4.184J/g.C)(T-23.6)
Solving for T gives 20.02 which is an answer choice (this is an MC question), but the correct answer is 20.3. What am I doing incorrectly?

[UG]

You forgot that the mass of the solution becomes 105 g + 9.40 g after the addition of KBr.

[Ben Cohen]

So the side of the equation that had the water would stay the same, but the side of the equation for KBr would have the mass changed?

[UG]

You will get (9.4g)(1mol/119g)(19900J/mol) = -(114.4g)(4.184J/g.C)(T-23.6)