Asked by Morgan
A 2.75 kg block of ice is heated until it completely melts. It is then heated again until it completely changes to water vapor. Is more energy required to melt the ice or to vaporize the liquid water.
Answers
Answered by
DrBob222
q1 to melt ice = mass ice x heat fusion.
q2 to raise T to 100 C = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2* to torn liquid H2O at 100 C to steam at 100 C = mass H2O x heat vaporization.
Total to melt ice = q1
Total to raise T from zero to 100 and turn steam is q2 + q2*
Compare q1 to (sum q2+q2*)
Note: The question is somewhat misleading. I assume the question means to heat the H2O from zero C to 100 C and continue heating until the liquid changes to vapor. But water has an appreciable vapor pressure at say room temperature and will vaporize completely if we wait long enough. The heat required to vaporize H2O at 25C is not the same as the sum of q2 and q2* above.
q2 to raise T to 100 C = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2* to torn liquid H2O at 100 C to steam at 100 C = mass H2O x heat vaporization.
Total to melt ice = q1
Total to raise T from zero to 100 and turn steam is q2 + q2*
Compare q1 to (sum q2+q2*)
Note: The question is somewhat misleading. I assume the question means to heat the H2O from zero C to 100 C and continue heating until the liquid changes to vapor. But water has an appreciable vapor pressure at say room temperature and will vaporize completely if we wait long enough. The heat required to vaporize H2O at 25C is not the same as the sum of q2 and q2* above.
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