Stanton D. answered • 04/23/14
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Meg,
The answer to your other question should set you on track to answer this one.
Stanton D.
Q (for 10g) = Sigma(all changes) = ((0.615 kg+0.524 kg)*21C*0.22 kcal/kgC)+ (2.00 kg * 21C * 1.00 kcal/kgC)
Scale up by factor of (100g/10g) and you should be there. I get ~473 kcal. (did you forget the scaleup?)
Note: Combustion is defined as producing products at standard conditions (i.e. liquid water and gaseous CO2): should you worry about that? Well, heats of combustion
should be ΔH, not ΔU (the latter also known as ΔΕ) -- i.e. at constant pressure -- but, depending on your course, you may not need to compensate for the ΔPV work done on the system when you "relax"
the bomb products to constant pressure (it's slightly positive, since a little O2 was consumed to make H2O, so that the surroundings actually adds a little work to the system when you replace that volume lost: that will
reduce slightly the net energy produced by the "system", hence reduce that 473 value a little -- I estimate by about 1 kcal [reasoning: CH2O makes H2O; 30g carbohydrate uses 1 mol O = 1/2 mol O2 ; then 100g carbohydrate
uses 10/6 mol O2; ΔΡV = nRT, you do the rest]; note that the O2 to CO2 transition is a "wash" -- there's no moles of gas change.)
I'll assume that if your course has covered stoichiometry of reactions, you can calculate the ΔPV correction from the formula for carbohydrate, if you need it, I've just given you a skeletal outline above.
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04/24/14
Leigh A.
I did scaleup, but I scaled up by 100 instead of 10 for some reason. Thank you so much!
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04/25/14
Leigh A.
04/24/14