Asked by Maggie
Find the total number of calories needed to change 23.0 g of ice at -34.0C to steam at 222.0C
Answers
Answered by
DrBob222
q1 = heat needed to raise T of ice at -34C to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tintial)
q2 = heat needed to melt ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion
q3 = heat needed to raise T of liquid water at zero C to liquid water at 100 C.
q3 = mass x specific heat H2O x (Tfinal-Tinitial)
q4 = heat needed to change liquid water at 100 C to steam(gas) at 100 C.
q4 = mass H2O x heat vaporization
q5 = heat needed to raise T of steam at 100 C to steam at 222 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial)
Total = q1 + q2 + q3 + q4 + q5.
q1 = mass ice x specific heat ice x (Tfinal-Tintial)
q2 = heat needed to melt ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion
q3 = heat needed to raise T of liquid water at zero C to liquid water at 100 C.
q3 = mass x specific heat H2O x (Tfinal-Tinitial)
q4 = heat needed to change liquid water at 100 C to steam(gas) at 100 C.
q4 = mass H2O x heat vaporization
q5 = heat needed to raise T of steam at 100 C to steam at 222 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial)
Total = q1 + q2 + q3 + q4 + q5.
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