Question

Can someone walk me through the steps on how to solve using these equations? Thank yoouuuu~~
1.11g of methane CH4 is burned in a bomb calorimeter containing 1000 grams of water. The initial temperature of water is 24.85oC. The specific heat of water is 4.184 J/g oC. The heat capacity of the calorimeter is 695 J/ oC. After the reaction the final temperature of the water is 35.65oC.

Using the formula water q =m•c• ΔT,calculate the heat absorbed by the water
Using the formula cal cal q =C • ΔT,calculate the heat absorbed by the
calorimeter.

Answer 1

I don't get it. You're given the formula(s) to use. All you need to do is to substitute the values in the problem.
For the water.
q = mc(Tfinal-Tinitial)
q = 1000g x 4.184 J/g*C x (35.65-24.85)= ?
q is + so heat is absorbed.

For the cal.
q = C*(Tfinal-Tinitial)
q = 695 J/C x (35.65-24.85) = ?
q is + so heat is absorbed.