A 100.0 mL of 0.500 M HBr at 20.29 oC is added to 100.0 mL of 0.500 M KOH (also at 20.29 oC). After mixing, the temperature rises to 23.65 oC. Calculate the heat of this reaction. [assuming the density and specific heat of HBr and KOH is the same as water, 1.0 g/mL; 4.18 J/g oC, and the volume of the solution is additive].

Answer : The heat of this reaction will be, 2.81 kJ

Explanation :

First we have to calculate the mass of water.

Given:

Density of water = 1.0 g/mL

Volume of water = 100.0 mL + 100.0 mL = 200.0 mL

Now we have to calculate the heat of this reaction.

Formula used :

where,

q = heat = ?

m = mass of water = 200.0 g

c = specific heat of water =

= final temperature =

= initial temperature =

Now put all the given values in the above formula, we get:

Thus, the heat of this reaction will be, 2.81 kJ

the answer is c i believe