Question goes like this:

When 25.0 ml of 1.0M H2SO4 is added to 50.0 ml of 1.0M NaOH at 25 CelSius in a calorimeter, the temperature of the aqueous solution increases to 33.9 Celsius. Assuming that the Specific heat of the solution is 4.18 J/(G*Celsius), that its density is 1.00 g/ml, and that the calorimeter itself absorbs a negligible amount of heat, calculate the [delta h=change in heat] in Kj for the reaction:

H2SO4 (aq) + 2 NaOH (aq) => 2 H2O(l) = Na2SO4

Problem:

Well, I compared my answer to the actual one which is [ -1.1 x 10^2] and I was off...

I utilized the equation q=c*m*delta T => [4.18]*[75g]*[33.9-25], managed to get 2.8*10^3

Any help would be appreciated.