[Illyaria]

When 1.500 g of solid potassium hydroxide is added to 25.00 mL of 2.010M hydrobromic acid at 22.5°C in a coffee-cup calorimeter, the temperature of the resulting solutions is measured and recorded for 180 s. The density of the hydrobromic acid is 1.159 g/mL, and the specific heat of the solution is 3.70 J/g·°C. Calculate the heat absorbed by the solution.

My attempt: q soln = (3.70J/g°C) {(25.00 mL HBr) (1.159g/ml) + 1.500 KOH) (22.5°C)

The part I am uncertain about is my ΔT at the end. The 22.5°C. I don't have a final temp to subtract the 22.5°C from to get ΔT unless there is something I'm not seeing?

[Borek]

The question is incomplete. ΔT should be calculated from the recorded data which is missing.

[Illyaria]

I thought so too. Thanks!