1. Given a liter of water just at boiling, how much energy does it take to boil it all?
2. If that liter of water was originally at 10°C, how much extra energy would it take to boil
the water?
Would it be:
1. Q=mL=(1kg)(2.26e6)=2.26e6
2. Q=mC(Tf-Ti)=(1kg)(4186J)(100C-10C)=376740J
2 answers
did not check arithmetic but that is the right idea
Goodness. Your use of units leaves much to be desired. Let units work for you.
a. energy=Hv*mass=2257kJ/kg*1liter*1kg/liter= 2257kJ
b. mcDeltaTemp=1kg*4.184kJ/kgC*90C=
= 90*4.184kJ
a. energy=Hv*mass=2257kJ/kg*1liter*1kg/liter= 2257kJ
b. mcDeltaTemp=1kg*4.184kJ/kgC*90C=
= 90*4.184kJ