Low or high enthalpy of combustion in relation to stability I know that this is a very basic question, but I am new to chemistry and I cannot figure it out on my own:

A combustion is always exothermic, so ΔH is negative. It says in my textbook that more stable molecules have a lower heat of combustion (I am assuming here that this is basically identical to stating that it has a lower enthalpy of combustion). 

What I do not understand is the following: more stable molecules have a lower enthalpy of formation than less stable molecules. When undergoing combustion, stable molecules should be expelling less heat than less stable ones. So shouldn't the enthalpy of combustion of more stable molecules be higher (i.e. less negative)?

N.B. The question is related to the following:
I was asked to determine which of the following two compounds has a lower enthalpy of combustion. I understand that B is more stable than A, but I am not able to translate that to "a lower enthalpy of combustion" because of my lack of understanding of the above…





Many thanks!
I have redrawn your two structures A and B in their most stable conformations. Note that A has all equatorial methyl groups while B has two equatorial groups and one that is axial. The estimated difference in the heats of formation and combustion is 1.8 kcal/mol, the same as the difference in energy between the equatorial and axial conformations of methyl cyclohexane. Because A and B produce the same amount of CO2 and H2O upon combustion, the more stable A's heat of combustion is 1.8 kcal/mol less than that for B. Compare your A and B with their enantiomers S and T at this ChemSE site at which time I addressed a similar question.