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Calculate the total heat (in J) needed to convert 22.00 g of ice at -6.00 degrees C to liquid water at .5 degrees C
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q=mcΔT where q=heat energy m=mass c=specific heat ΔT=change in temperature=Tf-Ti
copy paste from yahho, much better, now...
q=22.00gX-6.00 (I might be wrong at this part, I might of actually had to put the amount of temperature that you WANT it to be not what it already is)X.5+-6.00
-122,X-5.95
725.9 Joules
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This looks right to me!
well uh @Hyperchemist, at 0 C there is something called sp. latent heat of fusion to convert ice to water which i think u have not considered. which if considered to be 336 J/g would be 336x22 = 7392 J.
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