What quantity of heat is necessary to convert 50.0 g of ice at 0.0 C into steam 100,0 C? The heat of fusion is 80.0 cal/g, the heat of vaporization is 540 cal/g, and the specific heat of water is 1.00 cal/gC.

Answer 1

q1 = heat needed to melt ice at zero C to liquid water at zero C.
q1 = mass ice x heat fusion

q2 = heat needed to raise T from zero C to 100 C.
q2 = mass melted ice x specific heat water x (Tfinal-Tinitial)

q3 = heat needed to vaporize water at 100 to steam at 100.
q3 = mass water x heat vaporization.

Total Q = q1 + q2 + q3.