Asked by Patience
what amount of energy in calories would it take to bring 20 grams of solid ice at negative twelve degrees celsius to liquid water at 100 degrees celsius?
Answers
Answered by
Elena
Q = ΔU1 + ΔU2 + ΔU3 =
=c1•m •ΔT1 +λ•m + c2•m •ΔT2 =
=20•10^-3•(2.060•10^3• 12 + 335•10^3 + 4.183•10^3•100) =
=15560 J
(c1, c2 –specific heats of ice and water, λ –heat of fusion of ice)
=c1•m •ΔT1 +λ•m + c2•m •ΔT2 =
=20•10^-3•(2.060•10^3• 12 + 335•10^3 + 4.183•10^3•100) =
=15560 J
(c1, c2 –specific heats of ice and water, λ –heat of fusion of ice)
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