# Heat Capacity...

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#### MDwannabe7

##### Future Doctor
10+ Year Member
15+ Year Member
Calculate the amount of heat needed to bring 10 g of ice from -15degC to 110degC. (Heat of fusion = 80 cal/g; heat of vaporization = 540 cal/g. The heat capacities of both ice and steam vary with temperature; for this problem, use the estimate of 0.5 cal/gK for both.)

I realized that this problem needed to be broken up into the amount of heat to get ice at -15 to ice at 0; the amount of heat to get from ice at 0 to water at 0; the amount of heat to get water at 0 to water at 100; the amount of heat to get water at 100 to steam at 100; and the amount of heat to get steam at 100 to steam at 110. I did all of the calculations with M=10 g, c=0.5; 80; 0.5; 540; 0.5 respectively, and t= 15, 100, and 10 for the three equations where it comes into play.

I obtained 6825 cal for the answer, but when I looked at the solutions, found that I was 500 cals off due to the fact that c for water at 0 to water at 100 should be 1 cal/gK instead of 0.5 cal/gK. Can anyone explain to me why this is the case, please?

#### ThePandaFactor

##### Full Member
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15+ Year Member
0.5 is for ice and steam, 1 is for water (based on definition of calorie)

#### engineeredout

##### Full Member
15+ Year Member
The calorie is defined as the amount of heat required to raise the temperature of one g of water one degree celcius. Its fundamental to know that liquid water has a heat capacity of 1 calorie/g*K

#### MDwannabe7

##### Future Doctor
10+ Year Member
15+ Year Member
Well, I feel stupid... Thanks for the help - I'll be sure to remember that one now!

#### ecoli

##### Full Member
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15+ Year Member
yup - the very definition of a calorie is based around this fact, fyi.