n = number of moles of sample of octane = 0.07 mol
Q = energy absorbed by a sample of octane = 3.5 x 10³ J
c = molar heat capacity of octane = 254.0 J/K* mol
ΔT = increase in temperature of octane = ?
Heat absorbed is given as
Q = n c ΔT
inserting the values
3.5 x 10³ J = (0.07 mol) (254.0 J/K* mol) ΔT
ΔT = (3.5 x 10³ )/((0.07) (254.0))
ΔT = 196.85 K
hence increase in temperature comes out to be 196.85 K
