If 35.00g of burning octane causes a 13.7°C increase in the temperature of the water in a calorimeter, what volume of water was in it?

The answer given: V= 29.5L

First calculate the amount of heat released by 35 g of octane
That heat is gained by water and detla t is 13.7degree C
s for water = 4.18 J/g.deg C
q = m x 4.18  x 13.7
now q you need to calculate from octane combustion 

First figure the measure of warmth discharged by 35 g of octane
That warmth is picked up by water and delta t is 13.7degree
Cs for water = 4.18 J/g. deg C
q = m x 4.18 x 13.7
presently q you have to figure from octane burning

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