Answer:
5447.93 KJ
Explanation:
Given:
The number of moles of octane , n = 0.015
Mass of the water heated, m = 250 grams
Temperature change of the water, ΔT = 371.2 K - 293.0 K = 78.2 K
The specific heat capacity of the water, C = 4.18 J/k g
Now,
the heat acquired by the water, Q
Q = mCΔT
on substituting the values, we get
Q = 250 × 4.18 × 78.2
or
Q = 81.719 KJ
Now, the heat of combustion = Q / n = 81.719 / 0.015 = 5447.93 KJ
