Question
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. 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
Answers
damon
did not check arithmetic but that is the right idea
bobpursley
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