posted by anonymous on .
How many calories must be supplied to 59 g
of ice at −6 C to melt it and raise the temperature of the water to 61 C? The specific heat of ice is 0.49 kcal/kg · K and of water is 1 kcal/kg · K; the heat of fusion of water is 79.7 kcal/kg, and its heat of vaporization is 418 kcal/kg.
Answer in units of kcal
Use two formulas.
Within a phase:
q=mass x specific heat x (Tfinal-Tinitial)
mass x heat vap at boiling point
mass x heat fusion at melting point
at melting point of ice. To change solid ice at zero C to liquid at zero C.
q = 59 g ice x heat fusion - 59 x 80 cal/g = ?
From ice at zero to water 100 C
q = mass x specific heat x(Tfinal-Tinitial) = 59 g x 1 cal/g x (100-0) = ?
Then add all of the q amount to obtain total q the divide by 1000 to convert to Kcal.