[raney nickel]

A young man eats a delicious slice of Chicago style pizza containing 1050kJ of energy.  If he weighs 70 kg and the heat capacity of the body s 4.18J/gK, then how much water would he have to evaporate to keep his body temperature constant? Hvap H20 37C = 2402J/g

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Did I do this right?
1. Find temp increase for NRG from pizza for a closed system.  2. use inc in temp to find del H

DelU=nCp delT : delU=1050000J, Tf = ?, Ti = 37C
1050000J = (70000g)(4.18J/gK)(Tf-Ti) :
3.589 = Tf - 310.15 : Tf=313.74 = 40.59C

delH = Cp DelT:
delH = (4.18)(40.59-37)=15.01J:
delHvap/delH=(2402J/g)/(15.01J)=160 g H20 need to evaporate

Please let me know if I messed it up somewhere.  I'd really appreciate it.  Thank you

[Borek]

You don't need his final temp. It is enough to assume 1050 kJ was completely used for evaporation - then the energy of the guy doesn't change.

But the question is slightly nonsensical. 1050 kJ in pizza is stored in chemical bonds and it doesn't make him immediately hotter. He can use this energy to climb the stairs in which case part of 1050 kJ will be converted into his potential energy and he won't have to sweat it out.

[raney nickel]

Yes, that makes a lot more sense. 

So, what if we got some magic duct tape and taped up all of his holes and he became a closed system?   Could I then consider that the T would have to increase, because there is no transfer? 

[Borek]

So, what if we got some magic duct tape and taped up all of his holes and he became a closed system?   Could I then consider that the T would have to increase, because there is no transfer?

No, as long as the pizza is not burnt (metabolically). Lump of coal may have energy content of 100 MJ, but your room won't get hotter just because the coal sits on the floor.