Ice to Steam Problem – Heat Change Example Problem

The ice to steam problem is a classic heat energy homework problem. This will outline the steps necessary to complete this problem and follow up with a worked example problem.

The amount of heat needed to raise the temperature of a material is proportional to the mass or amount of the material and the magnitude of the temperature change.

The equation most commonly associated with the heat needed is

Q = mcΔT

where
Q = Heat energy
m = mass
c = specific heat
ΔT = change in temperature = (T_final – T_initial)

A good way to remember this formula is Q = “em cat”.

You may notice if the final temperature is lower than the initial temperature, the heat will be negative. This means as the material cools, energy is lost by the material.

This equation only applies if the material never changes phase as the temperature changes. Additional heat is required to change from a solid to a liquid and when a liquid is changed into a gas. These two heat values are known as the heat of fusion (solid ↔ liquid) and heat of vaporization (liquid ↔ gas). The formulas for these heats are

Q = m · ΔH_f
and
Q = m · ΔH_v

where
Q = Heat energy
m = mass
ΔH_f = heat of fusion
ΔH_v = heat of vaporization

The total heat is the sum of all the individual heat change steps.

Let’s put this in practice with this ice to steam problem.

Ice to Steam Problem

Question: How much heat is required to convert 200 grams of -25 °C ice into 150 °C steam?

Useful information:
Specific heat of ice = 2.06 J/g°C
Specific heat of water = 4.19 J/g°C
Specific heat of steam = 2.03 J/g°C
Heat of fusion of water ΔH_f = 334 J/g
Melting point of water = 0 °C
Heat of vaporization of water ΔH_v = 2257 J/g
Boiling point of water = 100 °C

Solution: Heating cold ice to hot steam requires five distinct steps:

1. Heat -25 °C ice to 0 °C ice
2. Melt 0 °C solid ice into 0 °C liquid water
3. Heat 0 °C water to 100 °C water
4. Boil 100 °C liquid water into 100 °C gaseous steam
5. Heat 100 °C steam to 150 °C steam

Step 1: Heat -25 °C ice to 0 °C ice.

The equation to use for this step is “em cat”

Q₁ = mcΔT

where
m = 200 grams
c = 2.06 J/g°C
T_initial = -25 °C
T_final = 0 °C

ΔT = (T_final – T_initial)
ΔT = (0 °C – (-25 °C))
ΔT = 25 °C

Q₁ = mcΔT
Q₁ = (200 g) · (2.06 J/g°C) · (25 °C)
Q₁ = 10300 J

Step 2: Melt 0 °C solid ice into 0 °C liquid water.

The equation to use is the Heat of Fusion heat equation:

Q₂ = m · ΔH_f

where
m = 200 grams
ΔH_f = 334 J/g

Q₂ = m · ΔH_f
Q₂ = 200 · 334 J/g
Q₂ = 66800 J

Step 3: Heat 0 °C water to 100 °C water.

The equation to use is “em cat” again.

Q₃ = mcΔT

where
m = 200 grams
c = 4.19 J/g°C
T_initial = 0 °C
T_final = 100 °C

ΔT = (T_final – T_initial)
ΔT = (100 °C – 0 °C)
ΔT = 100 °C

Q₃ = mcΔT
Q₃ = (200 g) · (4.19 J/g°C) · (100 °C)
Q₃ = 83800 J

Step 4: Boil 100 °C liquid water into 100 °C gaseous steam.

This time, the equation to use is the Heat of Vaporization heat equation:

Q₄ = m · ΔH_v

where
m = 200 grams
ΔH_v = 2257 J/g

Q₄ = m · ΔH_f
Q₄ = 200 · 2257 J/g
Q₄ = 451400 J

Step 5: Heat 100 °C steam to 150 °C steam

Once again, the “em cat” formula is the one to use.

Q₅ = mcΔT

where
m = 200 grams
c = 2.03 J/g°C
T_initial = 100 °C
T_final = 150 °C

ΔT = (T_final – T_initial)
ΔT = (150 °C – 100 °C)
ΔT = 50 °C

Q₅ = mcΔT
Q₅ = (200 g) · (2.03 J/g°C) · (50 °C)
Q₅ = 20300 J

Find the total heat

To find the total heat of this process, add all the individual parts together.

Q_total = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Q_total = 10300 J + 66800 J + 83800 J + 4514400 J + 20300 J
Q_total = 632600 J = 632.6 kJ

Answer: The heat needed to convert 200 grams of -25 °C ice into 150 °C steam is 632600 Joules or 632.6 kiloJoules.

The main point to remember with this type of problem is to use the “em cat” for the parts where no phase change occurs. Use the Heat of Fusion equation when changing from solid to liquid (liquid fuses into a solid). Use the Heat of Vaporization when changing from liquid to gas (liquid vaporizes).

Another point to keep in mind is the heat energies are negative when cooling. Heating a material means adding energy to the material. Cooling a material means the material loses energy. Be sure to watch your signs.

Heat and Energy Example Problems

If you need more example problems like this one, be sure to check out our other heat and energy example problems.

Specific Heat Example Problem
Heat of Fusion Example Problem
Heat of Vaporization Example Problem

Other Physics Example Problems
General Physics Worked Example Problems