# When burning the contents of two identical cans, one with oil and the other with kerosene

When burning the contents of two identical cans, one with oil and the other with kerosene, where will more heat be released?

Given:
V1 = V2 = V – the volume of cans with oil and kerosene is the same;
ro1 = ro2 = ro = 800 kg / m3 – density of oil and kerosene;
q1 = 41 * 106 Joule / kilogram – specific heat of combustion of oil;
q2 = 40.8 * 106 Joule / kilogram – specific heat of combustion of kerosene.
It is required to determine the combustion of which substance will release more heat.
When burning oil, heat will be released, equal to:
Q1 = q1 * m1 = q1 * V1 * ro1 = q1 * V * ro.
When kerosene burns, heat will be released, equal to:
Q1 = q2 * m2 = q2 * V2 * ro2 = q2 * V * ro.
Then:
Q1 / Q2 = (q1 * V * ro) / (q2 * V * ro) = q1 / q2 = 41 * 10 ^ 6 / (40.8 * 10 ^ 6) = 41 / 40.8 = 1.005.
Since Q1 / Q2 = 1.005> 1, more heat will be released during oil combustion.
Answer: more heat will be released when oil burns.

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