Question

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371.2 K. What is the enthalpy of combustion of octane? The specific heat capacity of water is 4.18 J/K g.

Answer 1

5447.93 KJ

Step-by-step 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