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October 25, 2014

October 25, 2014

Posted by **Tim** on Sunday, April 6, 2008 at 9:01am.

- Physics -
**Damon**, Sunday, April 6, 2008 at 11:00amLets do it for one kg and say it absorbs 4.1 * 10^5 Joules to make the numbers easier.

first raise temp of ice from -13 to 0

Joules = 2000 * 13 = 26,000 = .26*10^5

so we have 4.1 - .26 = 3.84*10^5 Joules left and 1 kg ice at zero C

Now melt the ice

Joules = 3.34 10^5 Joules/kg heat of fusion

so we have (3.84 -3.34) = .5*10^5 Joules left and water at zero deg C

now heat the water

Joules = 5*10^4 = .4190*10^4 * (T-0)

T = 11.9 deg C

- Physics -
**bobpursley**, Sunday, April 6, 2008 at 11:01amHow much heat is required to bring the block of ice up to 0C?

How much heat is required to melt the ice at 0C?

If there is heat left over, then final temp can be calculated by

Heatleftover=10kg*specificheatwater(Tf-0)

solve for Tf.

The statement "the block absorbs 4.10106 J of heat" makes no sense to me.

- Physics -
**Anonymous**, Monday, September 16, 2013 at 1:00pmcalories given off when 87g {\rm g} of water cools from 47 ∘ C ^\circ C to 24 ∘ C

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