[kauzs-17]

Hello, I'm stuck at a certain part of my question.
This is the original question: If all the energy obtained from burning 1.00 pound of propane is used to heat water, how many kg of water can be heated from 20.0°C to 45.0°C?

So, I converted lbs > g > moles of propane and multiplied it by the given ΔHcomb, which is -2219.9. However, the way my professor showed it done, the + amount of moles times the -ΔHcomb resulted in a positive q(heat). Why wouldn't the answer be negative?
I'm guessing it might be because when you burn propane you're getting energy as a product, but I didn't think you could just change the sign like that.

[Corribus]

The sign of the enthalpy just tells you whether heat is consumed or released by the reaction.  (The negative sign tells you it's being released, which shouldn't be unexpected for a combustion.)  Now find out how much water that amount of heat (absolute value) can change the water's temperature over the range specified.. 

[kauzs-17]

So basically qwater = -qpropane, correct? What leaves the propane (q<0) goes to the water (q>0)?

[Borek]

What leaves the propane (q<0) goes to the water (q>0)?

Exactly.