**Problem:**

What is the heat in Joules required to convert 25 grams of -10 °C ice into 150 °C steam?

Useful information:

heat of fusion of water = 334 J/g

heat of vaporization of water = 2257 J/g

specific heat of ice = 2.09 J/g·°C

specific heat of water = 4.18 J/g·°C

specific heat of steam = 2.09 J/g·°C

**Solution:**

The total energy required is the sum of the energy to heat the -10 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 150 °C.

**Step 1:**Heat required to raise the temperature of ice from -10 °C to 0 °C Use the formula

q = mcΔT

where

q = heat energy

m = mass

c = specific heat

ΔT = change in temperature

q = (25 g)x(2.09 J/g·°C)[(0 °C - -10 °C)]

q = (25 g)x(2.09 J/g·°C)x(10 °C)

q = 522.5 J

Heat required to raise the temperature of ice from -10 °C to 0 °C = 522.5 J

**Step 2:**Heat required to convert 0 °C ice to 0 °C water

Use the formula

q = m·ΔH

_{f}

where

q = heat energy

m = mass

ΔH

_{f}= heat of fusion

q = (25 g)x(334 J/g)

q = 8350 J

Heat required to convert 0 °C ice to 0 °C water = 8350 J

**Step 3:**Heat required to raise the temperature of 0 °C water to 100 °C water

q = mcΔT

q = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]

q = (25 g)x(4.18 J/g·°C)x(100 °C)

q = 10450 J

Heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J

**Step 4:**Heat required to convert 100 °C water to 100 °C steam

q = m·ΔH

_{v}

where

q = heat energy

m = mass

ΔH

_{v}= heat of vaporization

q = (25 g)x(2257 J/g)

q = 56425 J

Heat required to convert 100 °C water to 100 °C steam = 56425

**Step 5:**Heat required to convert 100 °C steam to 150 °C steam

q = mcΔT

q = (25 g)x(2.09 J/g·°C)[(150 °C - 100 °C)]

q = (25 g)x(2.09 J/g·°C)x(50 °C)

q = 2612.5 J

Heat required to convert 100 °C steam to 150 °C steam = 2612.5

**Step 6:**Find total heat energy

Heat

_{Total}= Heat

_{Step 1}+ Heat

_{Step 2}+ Heat

_{Step 3}+ Heat

_{Step 4}+ Heat

_{Step 5}

Heat

_{Total}= 522.5 J + 8350 J + 10450 J + 56425 J + 2612.5 J

Heat

_{Total}= 78360 J

**Answer:**

The heat required to convert 25 grams of -10 °C ice into 150 °C steam is 78360 J or 78.36 kJ.