# What quantity of heat is necessary to convert 50 g of ice at 0 degrees C into steam at 100 degrees C?

## The heat of fusion of ice is 80 cal/g., the heat of vaporization of water is 540 cal/g., and the specific heat of water is 1 cal/g.*degrees C.

Aug 5, 2016

The amount of heat required is 36 kcal.

There are three separate heats involved in this problem:

1. ${q}_{1}$ = heat required to melt the ice to water at 0 °C
2. ${q}_{2}$ = heat required to warm the water from 0 °C to 100 °C
3. ${q}_{3}$ = heat required to convert the water to steam at 100 °C

q = q_1 + q_2 + q_3 = mΔ_text(fus)H + mcΔT + mΔ_text(vap)H

q_1 = mΔ_"fus"H = 50 color(red)(cancel(color(black)("g"))) × 80color(white)(l) "cal"·color(red)(cancel(color(black)("g"^"-1"))) = "4 000 cal"

q_2 = mcΔT = 50 color(red)(cancel(color(black)("g"))) × 1 color(white)(l)"cal"·color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × 100 color(red)(cancel(color(black)("°C"))) = "5 000 cal"

q_3 = mΔ_"vap"H = 50 color(red)(cancel(color(black)("g"))) × 540color(white)(l) "cal"·color(red)(cancel(color(black)("g"^"-1"))) = "27 000 cal"

$q = {q}_{1} + {q}_{2} + {q}_{3} = \text{4 000 cal + 5 000 cal + 27 000 cal" = "36 000 cal" = "36 kcal}$