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Hi eveyone,
need some help in solving this.
How much heat must be added to bring 20g of ice at 0 celsius water vapor at 100 celsius ?
(The heat of fusion of ice is 80 cal/g; the heat of vaporization of water is 540 cal/g)
this is what I did, correct me if i am wrong.
( 20g*80*25) + (20*80) +(20* 4.186*100) +
raise temp of ice to 0 celsius; to melt the ice; raise temp 0 to 100;
(20*540)
turn water to vapor; answer 33144 cal = 33,144 kcal
Is that right the way I set it up?
Thanks
need some help in solving this.
How much heat must be added to bring 20g of ice at 0 celsius water vapor at 100 celsius ?
(The heat of fusion of ice is 80 cal/g; the heat of vaporization of water is 540 cal/g)
this is what I did, correct me if i am wrong.
( 20g*80*25) + (20*80) +(20* 4.186*100) +
raise temp of ice to 0 celsius; to melt the ice; raise temp 0 to 100;
(20*540)
turn water to vapor; answer 33144 cal = 33,144 kcal
Is that right the way I set it up?
