Consider the combustion of propane. Delta H is -2221 kJ. Assume all heat is from combustion of propane. What mass of propane must be burned in order to furnish this amount of energy assuming the heat transfer process is 60% efficient.
2221 kJ is dH. You lose 40%.
So 60% of what number is 2221. That must be about 3700 kJ which is what you need (that's just a close guess)
C3H8 + 5O2 ==>3O2 + 4H2O
molar mass C3H8 is 44 grams
You know 44 g C3H8 must be burned to produce about 2221 kJ.
So common sense says
44 g x (3700/2221) = ?
In other words, 44 g, if 100% would produce 2221 kJ BUT the process in the problem loses 40% and you must actually produce about 3700 kJ. So that takes more propane. The ratio 3700/2221 tells you how much more.posted by DrBob222