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3000 cal.
27000 cal.
135000 calories.
270000 calories

2007-10-04 05:16:28 · 2 answers · asked by thugluvn20 1 in Science & Mathematics Physics

2 answers

500 calories extracted from 1kg ice will drop the temperature by about 1C so multiply your temperature drop by 500

2007-10-04 05:29:55 · answer #1 · answered by Anonymous · 0 0

You need to supply an ending temperature. Lets call it T for now.

dQ = mcp*dT hwere dQ=heat supplied and dT = change in temperature

For your problem, the ice is getting colder so dT = 0 - T = -T. Then

dQ = -0.5* 1 *T calories

2007-10-04 12:21:56 · answer #2 · answered by nyphdinmd 7 · 0 1

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