When one mole of C6H6 is burned, 3.27 MJ of heat is produced. When the heat from burning 7.47 g of C6H6 is added to 5.69 kg of water at 21.0°C, what is the final temperature of the water?

Answer 1

How much heat do you get from the 7.47 g C6H6 if you get 3.27E6 J from 1 mol C6H6 (78 grams)? That's 3.27E6 J x (7.47/78) = ? = q

Then q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
I would convert kg H2O to g and use specific heat as 4.184 J/g*C and solve for Tfinal after substitution of the numbers.

Answer 2

Thank you DrBob222 I figured out the problem out thanks to your help.