Trying to figure how much heat is lost when 20 g of water at 45 degrees Celsius is changed to ice.

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Originally Posted by djw

Trying to figure how much heat is lost when 20 g of water at 45 degrees celcius is changed to ice.

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Using the given formula, I keep getting a different answer than the book gives.. not even close. Book gives answer of 10000 C.

You have to cool the ice from 45 degrees to 0 degrees Celsius. That's 1 cal/(g·°C).

Then, you have to subtract the "latent heat of fusion", 79.72 cal/g

If you have 20 grams, and 45 degrees go cool, then cooling requires 45 * 1 * 20 = 900 calories.

Heat of fusion is 79.72 cal/g * 45g = 3587.4 calories

Total is 4487.4 calories.

That's a factor of 2.2 smaller than the 10,000 calories that you say your book claims. I'd say the book is wrong unless there's something I'm missing.