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March 5, 2015

March 5, 2015

Posted by **Ceres** on Thursday, September 24, 2009 at 5:48pm.

To melt ice to 50g of water, I think that it will take:

50g(334J/g) = 167,000J

When cooling, the hot water should release:

Specific heat capacity of water(ΔT)(grams of water)(conversion to J from calories) = J released by cooling

(.5cal/g*oC)(80oC)(50g)(4.18J/cal) = 8360J

This is only about half of the energy that is needed to convert the ice to water, so I am missing something. Any ideas?

- Physics - ice to liquid -
**DrBob222**, Thursday, September 24, 2009 at 6:03pmI don't know why you converted J to calories. All of that seems unnecessary.

How much heat do you have in the 80 C water if it moves from 80 C to zero (which the problem states as the final T). Therefore,

q = mass water x specific heat water x delta T.

q = 50 x 4.18 x 80 = 16,720 Joules.

How much ice can you melt with that?

16,720/334 = ??

This is almost the same problem we worked on a couple of days ago with the block of ice.

- Physics - ice to liquid -
**DrBob222**, Thursday, September 24, 2009 at 6:15pmIf you want to work it with calories, the specific heat of water is 1.0 calorie/gram and the heat of fusion for ice is 80 cal/g, therefore

mass H2O x 1 cal/g x 80 = 4,000 calories.

4000 calories/80 cal/g = 50 g ice.

- Physics - ice to liquid -
**Ceres**, Thursday, September 24, 2009 at 9:42pmI am sorry to be a bother - I knew it was like the other problem but I couldn't figure out how. I am taking my class online and am struggling to learn the concepts, basically on my own. While I do well in math, this class has been very difficult for me to get my mind around - it's like learning a new language - with very little help. I now understand what you wrote - my one equation, 50g(334J/g) = 16,700J helped me to see how much energy I needed, I just didn't realize that the book was asking me to go back a chapter to get the equation I used last week. Now I know how to find where that energy comes from, and it even makes sense. Thanks for your patience and help. :-)

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