Molar Heat ... Please Help?

Question

The molar heat of combustion of propane is -2.2 x 10^3 KJ. How many grams of propane would have to be combusted to raise 1 kg of water from 22C to 100C. Assume no heat is absorbed by the container/air.

Please Explain How To Solve The Problem...

Answer 1

Let's use the 4.184 J/g-K constant for water to find the energy required to raise its temp by the 78 degrees:

dH = mCdT

dH = 1000g (4.184J/g-K)(78K)

dH = 326,352 joules

let's find the moles of propane required to supply that:

326.352 kJ @ 1 mole / 2.2e3 kJ = 0.148 moles of propane

let's find the mass of that amount of propane:

0.148 moles @ 44.1 g mol-1 = 6.54 grams

your answer is: 6.54 grams of propane