Question

Suppose a substance has a heat of fusion equal to 65 cal/g and a specific heat of 0.70 (cal/g*°C) in the liquid state. If 4.0 kcal of heat are applied to a 50-g sample of the substance at a temperature of 24°C, what will its new temperature be? What state will the sample be in? (melting point of the substance = 97°C; specific heat of the solid = 0.38 (cal/g*°C) ; boiling point of the substance = 760°C).

Answer 1

Calculate for the amount of heat needed to bring the solid to its melting point.
H1 = (50g)(0.38 cal/g°C)(97°C - 24°C)
= 1387 cal
Next, calculate for the heat of fusion,
H2 = (50 g)(65 cal/g) = 3250 cal

The sum of these two heats is 4637 cal. This means that the solid was brought to its melting point. However, the heat supplied was not enough to really melt the substance.