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? :scared: Thanks

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? :scared: Thanks
The problem says the ice is at 0 C.

Anyone wanna take a shot . Don't know if I am right or wrong.

There is something wrong with either your solution or the question.

Clarify what the temp starts at.

ok, let me rephrase the question .
How much heat is required to raise a 20 g of ice from -25 celcius to water vapor 100 celsius. Sorry, for the typo.:D

anyone knows how to solve this?

anyone?

anyone?
your kcal conversion is wrong but everything else looks good.