Heat Equations

These equations will allow you to calculate energy changes for H₂O. These changes may include phase changes, temperature changes, or a combination of the two. You need a table of heats in order to use these equations. The equations and the table of heats will always be given to you. You need to know when and how to use them.

Heat Equations

Type	Equation	When to use it	Variable names
Temperature Change	Q = m · ΔT · c	When temperature of H₂O changes within a single phase. For temp changes in more than one phase, use this equation multiple times.	Q = heat (kJ) m = mass (kg) ΔT = change of temperature ΔT = T_final - T_initial c = sp.ht = specific heat capacity
Phase Change S ←→ L	Q = m · HF	When ice changes to water or water changes to ice	Q = heat (kJ) m = mass (kg) HF = heat of fusion
Phase Change L ←→ G	Q = m · HV	When water changes to steam or steam changes to water	Q = heat (kJ) m = mass (kg) HV = heat of vaporization

Heat Value Table

Heat of Fusion	Heat of Vaporization	Specific Heat of Ice	Specific Heat of Water	Specific Heat of Steam
334 ^kJ/_kg	2260 ^kJ/_kg	2.095 ^kJ/_{kg °C}	4.19 ^kJ/_{kg °C}	2.0267 ^kJ/_{kg °C}

Steps for Solving Heat Equation Problems

Read the problem carefully to figure out the starting place

You need to know the starting temperature and the starting phase
If no temperature is given, then you need to infer from the information given

Determine the ending place

You need to know the ending temperature and phase
If no second temperature is given, then you probably just have a phase change to solve
Watch for phase differences (ice, water, steam)

Determine the number of steps you need to use

Use the heating curve to help you by mapping out the temperatures

If you go up or down a diagonal, you have a temperature change
If you go across a flat line, you have a phase change
Every part of the heating curve requires its own formula

Calculate each part of the heating curve using the equations and heat values above

Temperature changes use the ΔT equation

Be careful. If you go from ice below 0°C to water (or steam) above 0°C, then you need to use the temperature equation more than once.
You can only use the temperature change equation for one phase at a time
Make sure you are using the specific heat for the phase you are currently calculating
You should calculate ΔT separately to avoid errors

Calculate phase change with the corresponding heat of fusion or heat of vaporization as needed

For multi-step problems, add together the heats you calculated from each part

Use the idea that Q_total = Q₁ + Q₂ + ...

To report your final answer, you should use your judgment for rounding, and include the proper unit

Rule of thumb for rounding: keep three or four non-zero digits, but at least tenths place

Ex: 123.456 → 123.5 kJ
Ex: 0.23542 → 0.235 kJ

You should only round your FINAL ANSWER, not the intermediate steps
The proper unit for these problems will usually be kilojoules (kJ)

(Click here to practice with the heat equation calculator and random question generator)

Example Problems

1. How much heat is needed to change 2 kg of ice into water at 0°C?

This is only a phase change because temperature is not changing. Because it is going from solid to liquid, you need to use the heat of fusion.

Q = m · HF
Q =(2)(334)
Q = 668 kJ

2. How much heat is needed to change the temperature of 2.2 kg of water from 10°C to 40°C?

This is only a temperature change because you start and end with water. Because it is in the water phase, you need the specific heat for water. Note that the temperature is increasing, so ΔT should end up being positive.

ΔT = T_final - T_initial
ΔT = 40 - 10
ΔT = 30°C

Q = m · ΔT · c
Q =(2.2)(30)(4.19)
Q = 276.54 kJ

3. How much heat is needed to change the temperature of 3.6 kg of steam from 130°C to 110°C?

This is only a temperature change because you start and end with steam. Because it is in the steam phase, you need the specific heat for steam. Note that the temperature is decreasing, so ΔT should end up being negative. Therefore, Q will also be negative.

ΔT = T_final - T_initial
ΔT = 110 - 130
ΔT = -20°C

Q = m · ΔT · c
Q = (3.6)(-20)(2.0267)
Q = -145.9224 kJ
Q = -145.9 kJ

4. How much heat is needed to change the temperature of 0.6 kg of ice from -45°C to -17°C?

This is only a temperature change because you start and end with ice. Because it is in the ice phase, you need the specific heat for ice. Note that the temperature is increasing, so ΔT should end up being positive. In this problem, you have to be very careful of your integers and signs.

ΔT = T_final - T_initial
ΔT = -45 - -17
ΔT = 28°C
Note that this is a positive number.

Q = m · ΔT · c
Q = (0.6)(28)(2.095)
Q = 35.196 kJ
Q = 35.2 kJ

5. How much heat is needed to raise 1.3 kg of ice at 0°C to 30°C?

Immediately you should see there is a temperature change. But 0°C to 30°C is in the liquid phase, and the sample starts as ice. So, this is a two-step problem. You first need to melt the ice (phase change) and then heat up the resulting water (temperature change). Calculate both separately with their necessary equations and then add the values together to find the total heat.

1. Phase Change (S → L, need HF)	2. Temperature Change (0→30°C is for water)
Q₁ = m · HF Q₁ = (1.3)(334) Q₁ = 434.2 kJ	ΔT = T_final - T_initial ΔT = 30 - 0 ΔT = 30°C Q₂ = m · ΔT · c Q₂ = (1.3)(30)(4.19) Q₂ = 163.41kJ

Now that you have the phase change heat and the temperature change heat, add them together to get the final answer.

Q_total = Q₁ + Q₂
Q_total = 434.2 kJ + 163.41 kJ
Q_total = 597.61 kJ
Q_total = 597.6 kJ

6. How much heat is needed to change 2.3 kg of water from 75°C to 120°C?

Okay, this one is even tougher. You have to change the temperature in the water phase, then boil into steam, then raise the temperature to 120°C. This is a three-step problem. Once you know that, just take it one step at a time. Watch the values you need for each step. You'll be fine.

1. Temperature Change (75→100, water)	2. Phase Change (L→G, need HV)
ΔT = T_final - T_initial ΔT = 100 - 75 ΔT = 25°C Note that this temp change is only within the water phase. You can't do 125-75 because that's steam and water. Q₁ = m · ΔT · c = (2.3)(25)(4.19) = 240.925 kJ Remember, don't round anything until the end.	Q₂ = m · HV Q₂ = (2.3)(2260) Q₂ = 5198 kJ
3. Temperature Change (100→120, steam)	Total Heat
ΔT = T_final - T_initial ΔT = 120 - 100 ΔT = 20°C Note that this temp change is only within the steam phase. You can't do 125-75 because that's steam and water. Q₃ = m · ΔT · c Q₃ = (2.3)(20)(2.0267) Q₃ = 93.2282 kJ Remember, don't round anything until the end.	Q_total = Q₁ + Q₂ + Q₃ Q_total = 240.925 kJ + 5198 kJ + 93.2282 kJ Q_total = 5532.1532 kJ = 5532.2 kJ Here at last is your final answer. It would take 5532.2 kJ of heat to do this experiment.

The most complicated problem you'll probably see would be going from a subzero temperature in ice (like -25°C) up to a superheated value in steam (like 140°C). That would be a whopping 5-step problem!