Calorimetry help?

[PharmlyDoc]

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Calculate the amount of heat needed to bring 10 g of ice from -15 degrees C to 110 degrees C. 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/g*K for both.

The answer is supposed to be 7,325 cal.

 

[blueheron]

Add up your components separately:

Fusion 80x10= 800 cal
Vaporization 540x10=5400 cal
Heating up the water 100x10= 1000 cal
Heating up ice and steam 25x10x.5= 125 cal

Gives you your 7325

 

[PharmlyDoc]

I don't understand why the 25 degrees celsius is used for the ice and steam? And I know that water is converted to steam at 100 degrees celsius, but why is the heat capacity only applied to 25 degrees celsius and not the 100 degrees celsius?

 

[blueheron]

Heat cap is 1 for water. That is the definition of a calorie. Raises 1 gram of water 1 degree C. Use this for 0-100 C. -15 to 0 use the .5 and from 100-110 use .5. Hence a total of 25 degrees at .5 cal/gC.