Imagine you are cooking rice. You add water, place the pot on the stove, and wait. After a few minutes, the water begins to bubble, and steam rises. But here is a question: how much heat does it take to boil that water? How much more to turn it into steam? Now think about a car engine, a power station, or the Earth's climate. All of these involve measuring and managing heat energy.
In this chapter, we are going to learn how to measure heat — not just temperature, but the actual amount of thermal energy transferred. We will meet two powerful ideas: specific heat capacity, which tells us how much heat is needed to raise the temperature of a substance, and latent heat, which tells us how much heat is needed to change its state. These ideas are essential for understanding everything from cooking to climate science. Let us begin.
I once asked a class, "Which takes more heat: heating 1 kg of water from 20°C to 100°C, or boiling that same 1 kg of water into steam?" Most students guessed heating it to 100°C. They were amazed to learn that boiling takes about seven times more heat. That is the power of latent heat. It is one of the most surprising and important ideas in thermal physics.By the end of this chapter, you should be able to:
Before we go deeper, let us remind ourselves of the difference between heat and temperature.
Think of a cup of hot tea and a bathtub of warm water. The tea might be at 80°C, while the bath is at 40°C. The tea has a higher temperature. But the bath contains much more heat energy because it has far more water. Temperature tells you how hot something is; heat tells you how much thermal energy it contains or transfers.
In this chapter, we are concerned with quantity of heat — the actual amount of thermal energy transferred or stored.
This distinction is crucial. A spark from a firework has a very high temperature (over 1 000°C), but it contains very little heat energy because it has very little mass. A swimming pool at 25°C contains far more heat energy. Temperature and heat are not the same thing. Never confuse them.Heat capacity is the amount of heat required to raise the temperature of an object by 1°C (or 1 K).
Heat capacity: C = Q / ΔT
C = heat capacity (J/°C or J/K), Q = heat energy absorbed (J), ΔT = temperature change (°C or K)
Heat capacity depends on the mass and the material of the object. A large object has a larger heat capacity than a small one made of the same material.
Think of a large pot of water and a small cup of water. It takes much more heat to raise the temperature of the large pot by 1°C than the small cup. The large pot has a larger heat capacity.
| Object | Heat Capacity | Why |
|---|---|---|
| Small cup of water | Low | Small mass |
| Large pot of water | High | Large mass |
| Metal spoon | Low | Low specific heat capacity |
| Same mass of water | High | High specific heat capacity |
Specific heat capacity is the amount of heat required to raise the temperature of 1 kg of a substance by 1°C (or 1 K).
Specific heat capacity: c = Q / (m × ΔT)
c = specific heat capacity (J/kg°C), Q = heat energy (J), m = mass (kg), ΔT = temperature change (°C)
Rearranged for heat energy: Q = m × c × ΔT
The SI unit of specific heat capacity is the joule per kilogram per degree Celsius (J/kg°C) or joule per kilogram per kelvin (J/kg K).
Specific heat capacity tells us how much energy a substance can store per kilogram per degree. Substances with high specific heat capacity can absorb a lot of heat without their temperature rising much. Water has a very high specific heat capacity.
| Substance | Specific Heat Capacity (J/kg°C) |
|---|---|
| Water | 4 200 |
| Ethanol | 2 400 |
| Aluminium | 900 |
| Copper | 390 |
| Iron | 450 |
| Lead | 130 |
| Mercury | 140 |
Think of water as a thermal sponge. It can absorb a lot of heat without its temperature rising much. This is why it is used in car radiators, why coastal areas have milder climates than inland areas, and why water is such an effective coolant.
Water's high specific heat capacity is one of the reasons life exists on Earth. The oceans absorb enormous amounts of heat from the Sun during the day and release it at night, moderating the planet's temperature. Without this property, the Earth's climate would be much more extreme.The formula Q = mcΔT allows us to calculate the heat energy absorbed or released by a substance when its temperature changes.
How to use the formula:
Watch how I do this. Suppose we have 2 kg of water and we want to heat it from 20°C to 80°C. The specific heat capacity of water is 4 200 J/kg°C.
ΔT = 80 − 20 = 60°C
Q = 2 × 4 200 × 60 = 504 000 J = 504 kJ
So it takes 504 kJ of heat to raise the temperature of 2 kg of water by 60°C.
Notice that the temperature change (ΔT) is the same whether you use Celsius or Kelvin, because a change of 1°C equals a change of 1 K. The formula works with either, as long as you are consistent. But always use kilograms for mass, not grams. Convert grams to kilograms by dividing by 1 000.When a substance changes state — from solid to liquid, or liquid to gas — it absorbs heat without its temperature changing. This heat is called latent heat.
Latent heat of fusion: The heat required to change 1 kg of a solid into liquid at its melting point without a change in temperature.
Latent heat of vaporisation: The heat required to change 1 kg of a liquid into gas at its boiling point without a change in temperature.
Latent heat: Q = m × L
Q = heat energy (J), m = mass (kg), L = specific latent heat (J/kg)
Specific latent heat of fusion of ice: 3.34 × 10⁵ J/kg
Specific latent heat of vaporisation of water: 2.26 × 10⁶ J/kg
Think of ice melting. As you add heat, the ice does not get warmer — it stays at 0°C until it has all melted. The heat is being used to break the bonds between the water molecules, not to raise the temperature. That is latent heat.
| Substance | Latent Heat of Fusion (J/kg) | Latent Heat of Vaporisation (J/kg) |
|---|---|---|
| Water | 3.34 × 10⁵ | 2.26 × 10⁶ |
| Ethanol | 1.04 × 10⁵ | 8.54 × 10⁵ |
| Lead | 2.45 × 10⁴ | 8.70 × 10⁵ |
| Mercury | 1.14 × 10⁴ | 2.96 × 10⁵ |
Heating and cooling curves show how the temperature of a substance changes as heat is added or removed. They reveal the role of latent heat.
Heating curve:
Cooling curve:
| Section | What Is Happening | Energy Used For |
|---|---|---|
| Sloped (heating) | Temperature rising | Increasing kinetic energy of particles |
| Flat (melting) | Solid → Liquid | Breaking bonds (latent heat of fusion) |
| Sloped (heating) | Temperature rising | Increasing kinetic energy of particles |
| Flat (boiling) | Liquid → Gas | Breaking bonds (latent heat of vaporisation) |
Think of a heating curve as a story of energy. Each flat section is a chapter where the energy is being used for something other than raising temperature — breaking the bonds that hold particles together.
Heating curves are a beautiful way to visualise the energy changes in a substance. They show clearly that during a change of state, the temperature does not change even though heat is being added. This is one of the most important ideas in thermal physics.When two substances at different temperatures are mixed, heat flows from the hotter one to the colder one until they reach the same final temperature. The heat lost by the hot substance equals the heat gained by the cold substance, assuming no heat is lost to the surroundings.
Principle of mixtures: Heat lost by hot object = Heat gained by cold object
m₁ × c₁ × ΔT₁ = m₂ × c₂ × ΔT₂
This principle is used to determine specific heat capacities and to solve problems involving mixtures.
Example: If you mix hot water with cold water, the final temperature will be somewhere between the two. If the hot water is 80°C and the cold water is 20°C, the final temperature might be 50°C, depending on the masses.
The principle of mixtures is a direct consequence of the law of conservation of energy. Heat is not lost — it is transferred from the hot object to the cold one. When you mix hot and cold water, the hot water cools down and the cold water warms up until they reach thermal equilibrium. This is physics in action.One of the most common experiments in thermal physics is determining the specific heat capacity of a substance. Here is how it is done:
Method:
Finding specific heat capacity: c_substance = (m_water × c_water × ΔT_water) / (m_substance × ΔT_substance)
Precautions:
| Application | How It Uses Heat Quantities |
|---|---|
| Car radiators | Water's high specific heat capacity absorbs engine heat |
| Cooling systems | Water or coolant absorbs heat from engines |
| Cooking | Heat is transferred from stove to food |
| Refrigeration | Latent heat is used to absorb heat from inside the fridge |
| Air conditioning | Latent heat of vaporisation cools the air |
| Weather and climate | Oceans moderate temperature due to high specific heat capacity |
| Power stations | Steam (latent heat) drives turbines |
| Solar water heaters | Water absorbs heat from the Sun |
Understanding quantity of heat also helps us understand insulation. Insulation reduces the rate of heat transfer, keeping hot things hot and cold things cold.
Types of insulation:
Think of a thermos flask. It is designed to minimise all three methods of heat transfer. It keeps your tea hot for hours. That is the power of insulation — and it all comes from understanding heat quantities.
Insulation is about slowing down heat transfer, not stopping it. No insulator is perfect. Heat will always find a way to flow from hot to cold. But good insulation can slow it down dramatically. This is why a thermos flask can keep a drink hot for hours.Heat energy (temperature change): Q = m × c × ΔT
Q = heat energy (J), m = mass (kg), c = specific heat capacity (J/kg°C), ΔT = temperature change (°C)
Heat energy (change of state): Q = m × L
L = specific latent heat (J/kg)
Specific heat capacity of water: 4 200 J/kg°C
Specific latent heat of fusion of ice: 3.34 × 10⁵ J/kg
Specific latent heat of vaporisation of water: 2.26 × 10⁶ J/kg
Principle of mixtures: Heat lost = Heat gained
m₁ × c₁ × ΔT₁ = m₂ × c₂ × ΔT₂
Heat capacity: C = Q / ΔT
Question: Calculate the heat energy needed to raise the temperature of 3 kg of water from 20°C to 70°C. The specific heat capacity of water is 4 200 J/kg°C.
Answer:
Formula: Q = m × c × ΔT
ΔT = 70 − 20 = 50°C
Substitution: Q = 3 kg × 4 200 J/kg°C × 50°C
Answer: Q = 630 000 J = 630 kJ
Units check: kg × J/kg°C × °C = J. The answer is in joules, correct for heat energy. Does 630 kJ make sense? Yes, it is enough energy to boil about 1.5 kg of water from room temperature.Question: Calculate the heat energy needed to melt 0.5 kg of ice at 0°C into water at 0°C. The specific latent heat of fusion of ice is 3.34 × 10⁵ J/kg.
Answer:
Formula: Q = m × L
Substitution: Q = 0.5 kg × 3.34 × 10⁵ J/kg
Answer: Q = 1.67 × 10⁵ J = 167 kJ
Units check: kg × J/kg = J. The answer is in joules. Notice that this is the heat needed just to melt the ice, without changing its temperature. Latent heat is the energy needed to break the bonds between the ice molecules.Question: 2 kg of water at 80°C is mixed with 3 kg of water at 20°C. Calculate the final temperature of the mixture. Assume no heat is lost to the surroundings. Specific heat capacity of water = 4 200 J/kg°C.
Answer:
Let the final temperature be T.
Heat lost by hot water = m₁ × c × (80 − T) = 2 × 4 200 × (80 − T)
Heat gained by cold water = m₂ × c × (T − 20) = 3 × 4 200 × (T − 20)
By the principle of mixtures: 2 × 4 200 × (80 − T) = 3 × 4 200 × (T − 20)
Divide both sides by 4 200: 2(80 − T) = 3(T − 20)
160 − 2T = 3T − 60
220 = 5T
T = 44°C
Answer: The final temperature is 44°C.
Units check: the answer is in °C. Notice that the final temperature is closer to the temperature of the larger mass (3 kg of cold water). This makes sense — the larger mass has a greater influence on the final temperature.Question: Calculate the total heat energy needed to convert 1 kg of ice at −10°C into water at 20°C. Specific heat capacity of ice = 2 100 J/kg°C, specific latent heat of fusion of ice = 3.34 × 10⁵ J/kg, specific heat capacity of water = 4 200 J/kg°C.
Answer:
Step 1: Heat ice from −10°C to 0°C: Q₁ = m × c_ice × ΔT = 1 × 2 100 × 10 = 21 000 J
Step 2: Melt ice at 0°C: Q₂ = m × L = 1 × 3.34 × 10⁵ = 334 000 J
Step 3: Heat water from 0°C to 20°C: Q₃ = m × c_water × ΔT = 1 × 4 200 × 20 = 84 000 J
Total heat: Q_total = Q₁ + Q₂ + Q₃ = 21 000 + 334 000 + 84 000 = 439 000 J = 439 kJ
Units check: the answer is in joules. Notice that the largest contribution comes from melting the ice (334 kJ). This is why latent heat is so important — it takes more energy to change state than to change temperature over a similar range.Question: A 0.5 kg metal block is heated to 100°C and then placed in 0.2 kg of water at 20°C. The final temperature of the mixture is 30°C. Calculate the specific heat capacity of the metal. Specific heat capacity of water = 4 200 J/kg°C.
Answer:
Heat gained by water: Q_water = m_water × c_water × ΔT_water = 0.2 × 4 200 × (30 − 20) = 0.2 × 4 200 × 10 = 8 400 J
Heat lost by metal: Q_metal = m_metal × c_metal × ΔT_metal = 0.5 × c_metal × (100 − 30) = 0.5 × c_metal × 70 = 35 × c_metal
By the principle of mixtures: 35 × c_metal = 8 400
c_metal = 8 400 / 35 = 240 J/kg°C
Answer: The specific heat capacity of the metal is 240 J/kg°C.
Units check: the answer is in J/kg°C. Does 240 J/kg°C make sense? Yes, it is lower than water (4 200) and copper (390), so it could be a metal like lead or a mixture. This experiment is a classic way to determine the specific heat capacity of an unknown metal.Graphs are essential for understanding quantity of heat.
Heating curve: If you plot temperature on the vertical axis and heat energy added on the horizontal axis, you get a heating curve.
Cooling curve: If you plot temperature on the vertical axis and time on the horizontal axis as a substance cools, you get a cooling curve.
Interpreting the flat sections:
These graphs help us visualise the energy changes in a substance. They turn abstract equations into pictures.
Think of a graph as a map of energy. The shape tells you how the substance responds to heat.
Apparatus: A metal block (e.g., copper or aluminium), a balance, a beaker of water, a heater, a thermometer, a calorimeter, and a stopwatch.
Method:
Calculation: If the metal block has mass 0.2 kg, the water has mass 0.1 kg, the initial water temperature is 20°C, the final temperature is 25°C, and the metal was heated to 100°C:
Q_water = 0.1 × 4 200 × 5 = 2 100 J
ΔT_metal = 100 − 25 = 75°C
c_metal = 2 100 / (0.2 × 75) = 2 100 / 15 = 140 J/kg°C
Precautions:
Likely source of error: Heat loss to the surroundings means the calculated specific heat capacity will be slightly lower than the true value. Use insulation and work quickly.
This experiment is a beautiful way to determine the specific heat capacity of a metal. It uses the principle of mixtures and the conservation of energy. Once you have done it, you will understand why metals heat up and cool down at different rates. It is a classic experiment in thermal physics.Specific heat capacity is the amount of heat required to raise the temperature of 1 kg of a substance by 1°C. Its SI unit is J/kg°C.
Q = m × c × ΔT.
The heat required to change 1 kg of a solid into liquid at its melting point without a change in temperature.
Q = m × L.
Q = m × c × ΔT = 2 × 4 200 × 30 = 252 000 J = 252 kJ.
Q = m × L = 0.2 × 3.34 × 10⁵ = 66 800 J = 66.8 kJ.
The heat energy is used to break the bonds between particles, not to increase their kinetic energy. So the temperature does not change.
Heat lost by the hot object equals the heat gained by the cold object, assuming no heat is lost to the surroundings.
(a) Define specific heat capacity and state its SI unit. [2]
(b) Calculate the heat energy needed to raise the temperature of 2.5 kg of water from 15°C to 75°C. Specific heat capacity of water = 4 200 J/kg°C. [3]
(c) Explain why water is used as a coolant in car radiators. [3]
(a) Specific heat capacity is the heat required to raise the temperature of 1 kg of a substance by 1°C. SI unit: J/kg°C. (2 marks.)
(b) ΔT = 75 − 15 = 60°C. Q = m × c × ΔT = 2.5 × 4 200 × 60 = 630 000 J = 630 kJ. (3 marks.)
(c) Water has a high specific heat capacity, so it can absorb a lot of heat with only a small rise in temperature. This makes it an effective coolant. (3 marks.)
(a) Define specific latent heat of vaporisation. [2]
(b) Calculate the heat energy needed to convert 0.5 kg of water at 100°C into steam at 100°C. Specific latent heat of vaporisation of water = 2.26 × 10⁶ J/kg. [3]
(c) Explain why steam burns are more severe than burns from boiling water. [4]
(a) The heat required to change 1 kg of a liquid into gas at its boiling point without a change in temperature. (2 marks.)
(b) Q = m × L = 0.5 × 2.26 × 10⁶ = 1.13 × 10⁶ J = 1.13 MJ. (3 marks.)
(c) Steam releases its latent heat of vaporisation when it condenses on the skin. This is a large amount of energy (2.26 × 10⁶ J/kg), much more than the energy released by boiling water cooling down. So steam causes more severe burns. (4 marks.)
(a) Describe an experiment to determine the specific heat capacity of a metal block. [5]
(b) A 0.3 kg metal block at 100°C is placed in 0.15 kg of water at 20°C. The final temperature is 28°C. Calculate the specific heat capacity of the metal. Specific heat capacity of water = 4 200 J/kg°C. [5]
(a) Measure mass of metal block. Heat it in boiling water to 100°C. Measure mass of water in calorimeter and record initial temperature. Transfer hot metal to calorimeter. Stir and record maximum temperature. Calculate c_metal using the principle of mixtures. (5 marks.)
(b) Q_water = 0.15 × 4 200 × 8 = 5 040 J. ΔT_metal = 100 − 28 = 72°C. c_metal = 5 040 / (0.3 × 72) = 5 040 / 21.6 = 233.3 J/kg°C. (5 marks.)
Explain why the temperature of a substance does not change during a change of state, even though heat is being added.
During a change of state, the heat energy is used to overcome the forces of attraction between the particles, not to increase their kinetic energy. Since temperature is a measure of the average kinetic energy of the particles, it does not change. The energy is stored as potential energy in the particles, not kinetic energy.
A student says, "Boiling water and steam at 100°C have the same heat content." Is this correct? Explain.
No. Steam at 100°C contains more heat energy than boiling water at 100°C because steam has absorbed the latent heat of vaporisation. This extra energy is released when the steam condenses. This is why steam burns are more severe than burns from boiling water.
Explain why coastal areas have milder climates than inland areas, using the concept of specific heat capacity.
Water has a high specific heat capacity, so it heats up and cools down slowly. In coastal areas, the ocean absorbs heat from the Sun during the day without a large temperature rise, and releases it at night without a large temperature drop. This moderates the climate, making coastal areas cooler in summer and warmer in winter compared to inland areas.
Heat is the currency of energy in the universe — and understanding how to measure it unlocks the secrets of everything from cooking to climate.
