Asked by Brad
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
Answered by
damon
did not check arithmetic but that is the right idea
Answered by
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.