Interactive textbook: Heat capacity and latent heat

Heat capacity

When the internal energy of a body is increased, either by doing work or transferring heat, most of the time the temperature rises. The rise in temperature is proportional to the amount of heat added and the constant of proportionality is called the heat capacity. It is dependant on the mass and material the body is made of:

Heat capacity is the amount of heat required to raise the temperature of a body by \(1\text{ K}\).

\(C={Q\over\Delta T}\)

  • \(C\) is heat capacity in \(\text{J K}^{-1}\)
  • \(Q\) is heat added in \(\text{J}\)
  • \(\Delta T\) is temperature increase in \(\text{K}\)

The heat capacity of a stainless steel kettle is about \(400\text{ J K}^{-1}\) and for a 5 kg stone \(5000\text{ J kg}^{-1}\). When a kettle of water is heated from \(20\text{ }℃\) to \(100\text{ }℃\),  \(3200\text{ J}\) of energy is used to heat the kettle. This is wasted energy since you want to heat the water (and not the kettle!). On the other hand, if a fireplace is surrounded by 50 \(5\text{ kg}\) stones that are heated from \(20\text{ }℃\) to \(60\text{ }℃\), they will store \(10^6\text{ J}\) of energy, enough to keep the room warm for several hours after the fire goes out. Stones are particularly good for this purpose as they conduct heat quite badly meaning that they lose heat slowly.

Specific heat capacity

Heat capacity relates to a body. Specific heat capacity relates the temperature change to heat transferred for a specific material:

Specific heat capacity is the amount of heat required to raise the temperature of \(\mathbf{1}\textbf{ kg}\) of material by \(1\text{ K}\).

\(c={Q\over m\Delta T}\)

  • \(c\) is specific heat capacity in \(\text{J kg}^{-1}\text{ K}^{-1}\)
  • \(m\) is mass in \(\text{kg}\)

For reference, the specific heat capacity of water is approximately \(4200\text{ J kg}^{-1}\text{ K}^{-1}\).

To heat \(1\text{ kg}\) of water from \(20\text{ }℃\) to \(100\text{ }℃\) requires \(3.36\text{ kJ}\) of energy. Water has quite a high specific heat capacity which, combined with the fact that it is cheap and readily available, makes it a very useful liquid to use for transferring heat energy from one place to another, both for cooling as in a car engine or heating as in domestic central heating systems.

To heat up a metal pan of water from 20℃ to 30℃ requires 3000 J of heat.

If the mass of the pan plus water is 1 kg what can you deduce about the specific heat capacity of the metal

If it was the same as water the heat require would be 4200 J




Heating and power

Electric kettles and other water heaters often have a power rating. Power is energy per unit time so we could rewrite the equation:

\(P={Q\over \Delta t}={mc\Delta T\over \Delta t}=mc{\Delta T\over\Delta t}\)

  • \(P\) is power in \(\text{W}\)
  • \(\Delta t\) is time taken in \(\text{s}\)

Increased power will give a faster increase in temperature.


An electric kettle has a power rating of 2.1 kW.

How long will it take to raise the temperature of 1 ltr of water from 20℃ to 100℃

P = mcΔT/t

t = 1 x 4200 x 80/2100


Measuring specific heat capacity

Mechanical energy to heat

To find the specific heat capacity of a material we need to measure the increase in temperature of a known mass when a known amount of energy is given to it. The most obvious way to do this is by doing work against friction. 'PASCO scientific' make a piece of apparatus designed to do just that.

The body being given energy is the aluminium cylinder with the rope wrapped round it. A mass is hung on the rope and the handle turned so the mass is lifted off the ground. The rope is not tied to the cylinder and so it starts to slip. If the handle is turned at a steady rate, the rope slips at just the right speed so that the mass doesn't move up or down. The friction on the rope then equals the weight of the mass.

In this animation, the tension on the right hand side of the rope is zero and the hanging weight is being held up by the friction between the rope and the cylinder. The person turning the handle is doing work against the friction.

The point of application of the friction is on the edge of the cylinder. Each revolution moves a distance equal to the circumference (\(2πr\)). The work done per revolution is equal to \(2πrF\).

But the friction is equal to \(Mg\) where \(M\) is the hanging mass, so the work done per revolution is \(2πrMg\).

If it takes \(n\) revolutions to raise the temperature of the cylinder (mass \(m\)) by \(ΔT\) then, assuming all the work is converted to internal energy:

\(2πrMgn = mcΔT\)

The specific heat capacity of aluminium is therefore:

\(c={2\pi rMg\over m\Delta T}\)

The experiment above is carried out with one large mass, M hanging from the two ropes.

The work done against friction per revolution is:

The friction = the difference between the two tensions.


Method of mixtures

Once we have determine the specific heat capacity of one material we can use it to find the specific heat capacity of others. By heating aluminium mechanically we find its specific heat capacity is \(900\text{ J kg}^{-1}\text{ K}^{-1}\). To heat \(1\text{ kg}\) from \(20\text{ }℃\) to \(100\text{ }℃\) would require \(72\text{ kJ}\) of energy. The method of mixtures uses the energy stored in one body to heat another.

Let's take a \(1\text{ kg}\) block of aluminium at \(100\text{ }℃\) and put it into \(1\text{ kg}\) of water at \(20\text{ }℃\). Heat flows from the hot aluminium to the cold water until thermal equilibrium is reached at \(34.12\text{ }℃\).

\(\text{heat lost by aluminium} = \text{heat gained by water}\)


\(c_w={59,292\over14.12}=4200\text{ J kg}^{-1}\text{ K}^{-1}\)

A block of metal with heat capacity 420 JK-1 is heated on a flame and then dropped into 1 ltr of water raising the temperature of the water by 50℃.

Calculate the temperature of the block.

420 x (T - 50) = 4200 x 1 x 50

T - 50 = 500

T = 450


Electric heating

By far the easiest way to measure specific heat capacity of a material is to heat a known mass using an electric heater such as a water boiler. We need to know the boiler's power. This is usually written on the bottom of the appliance but we could alternatively measure the rate at which it heats water.

\(P={Q\over\Delta t}=mc{\Delta T\over\Delta t}\)

A water boiler with a power rating of \(2\text{ kW}\) will heat \(1\text{ kg}\) of water at a rate of \(0.5\text{ K s}^{-1}\).


A metal block is placed in a 2 kW electric water boiler together with 1 ltr of water.

In 100 s the temperature is raised by 40℃, what is the heat capacity of the block?

2000 x 100 = 1 x 4200 x 40 + c x 40

200000 - 168000 = 40c

32000 = 40c



Change of state

The temperature doesn't always change when heat is added. One reason could be because heat is being lost to the surroundings as as fast as it is being added. Another possibility is that the material is changing state.

You will have experienced this when boiling a kettle. When you switch it on the temperature rises, but the rate of temperature rise decreases as the water gets hotter. This is because more heat is lost at higher temperatures. When the water reaches \(100\text{ }℃\) the temperature stops rising and instead turns to steam.

The temperature is constant so there can be no change in kinetic energy. Instead, the heat increases the potential energies of the molecules and thus changing their position. Note that this doesn't necessarily mean the molecules are moving apart.

Latent heat of vaporisation

The latent heat of vaporisation is the amount of heat required to change the state of one kilogram of a substance from liquid to gas without change in temperature:

\(L_\text{v}={Q\over m}\)

  • \(L_\text{v}\) is latent heat of vaporisation in \(\text{J}\)
  • \(Q\) is heat added in \(\text{J}\)
  • \(m\) is the mass of substance in \(\text{kg}\)

The latent heat of vaporisation of water can be measured using the simple kettle experiment. This time it's best not to use the power rating of the kettle because so much heat is lost when the kettle is boiling. Instead, the gradient of the temperature vs time graph can be used to calculate the amount of heat added to the water per second just before it boils by using the equation:

\(P={Q\over\Delta t}=mc{\Delta T\over\Delta t}\)

Once you know the energy given to the water per second, the latent heat can be found by measuring the mass of water turned to steam in some time \(Δt_\text{s}\). The heat given to the water in this time is \(PΔt_\text{s}\) so the latent heat is:

\(L_\text{v} = {PΔt_\text{s}\over m}\)

If you do this experiment you will get a value of around \(2.2\text{ MJ kg}^{-1}\). This is a high value but is to be expected; it takes only a few minutes to raise the temperature to boiling point but more like half an hour to boil it dry (something you shouldn't try!).

When doing this experiment the value gained is often bigger than expected. This could be due to two reasons:

  • More heat is lost per second than just before boiling so the measured power is bigger than the heat given to the water.
  • Some steam condenses and falls back into the kettle so the mass of steam measured is less than the actual mass turned to steam.

A metal block with heat capacity of 500 J K-1 is dropped into 1 ltr of water at 20℃ causing 10 dl to turn to steam.

What was the temperature of the block?

heat require to raise temp of water from 20℃ to 100℃ + heat required to turn water into steam = heat from block

1 x 4200 x 80 + 0.1 x 2.2 x 106 = 500 x (T - 100)

556000 = 500(T - 100)

Latent heat of fusion

The latent heat of fusion is the amount of heat required to change the state of one kilogram of a substance from solid to liquid without change in temperature:

\(L_\text{f}={Q\over m}\)

  • \(L_\text{f}\) is latent heat of fusion in \(\text{J}\)

Again, the latent heat of water can be measured with a simple experiment - this time based on the method of mixtures. If ice cubes at \(0\text{ }℃\) are put into a jug of warm water at temperature \(T_1\), the water cools to a final temperature \(T_2\). Heat from the water first melted the ice then raised its temperature to \(T_2\).

\(\text{heat from warm water} = \text{heat used to melt ice} + \text{heat used to warm melted ice to }T_2\)

\(m_\text{w}c(T_1 - T_2) = m_\text{i}L_\text{f} + m_i c(T_2 - 0)\)

Things to look out for:

  • All the ice must melt.
  • The final temperature must be recorded as soon as the last bit of ice melts.

Heat transfer

In all of the examples above we have ignored the fact that not all of the heat is transferred to the body we are interested in. Some is lost. In the first example we should say:

\(W = mcΔT + \text{heat lost}\)

\(c={W-\text{heat lost}\over m\Delta T}\)

If we ignore the heat lost we will get a value for specific heat capacity that is too big. To correct this we must either find a way to calculate the heat lost or minimise it. To achieve the latter, we need to understand the way heat is transferred.


Conduction is the way heat is transferred by molecules passing kinetic energy from one to another. A poor conductor of heat is called an insulator. In the simulation below, springs represent the interatomic force.

A good conductor will be a material with strong interatomic forces. Metals are particularly good conductors because the atoms in a metal share delocalised electrons that help to pass on the kinetic energy. Gases are poor conductors since there are no interatomic forces; kinetic energy is only transferred when the atoms collide.

To reduce heat loss by conduction we can cover the apparatus with an insulating material.


Convection is the way heat is transferred in fluids (liquids and gases). In this process large quantities of the hot fluid moves about. If a volume of air is heated the atoms move faster, pushing aside the surrounding atoms.

This causes a reduction in the density of this volume of gas so, according to Archimedes' principle, it will experience a buoyant force greater than its weight resulting in upward movement. If this gas comes into contact with a solid object the fast moving atoms transfer energy to the solid. Heat is transferred by the gas moving in convection currents.

Gases are good insulators but convection enables heat to be transferred through them. To prevent heat loss by convection, a lid can be used or gas can be trapped in small "pockets" to utilise its insulating properties. This is the principle of most insulating material such as foam, wool and down.

Removing the air completely prevents both conduction and convection. A vacuum flask is made of two layers of glass or stainless steel with a vacuum in between, keeping coffee hot for a long time!


There is no air between the Sun and the Earth but heat is still transferred between them. This transfer type is called radiation. Heat radiation is similar to light waves but has a wavelength a bit longer than red: "infrared". Infrared radiation is invisible to the human eye but can be detected with an infrared camera. The amount of radiation emitted by a body is proportional to the fourth power of its temperature in Kelvin and also to its colour. The process of radiation is similar to absorption so a black body that absorbs most radiation also emits most. Silver coloured bodies don't absorb or radiate effectively; they reflect radiation.

To reduce radiation from a body we can paint the body silver or cover it with aluminium foil.

You will learn more about the relationship between electromagnetic radiation and matter in the Energy Production section of the course.

Putting a lid on a pan of water reduces the time taken to boil.

It reduces the heat lost by convection.

The image show a radiator.

Hot water flows through the radiator giving heat to the room.

The best colour for the radiator is:

Only a small amount of heat is actually radiated from the radiator, it actually heats the room by convection.

The image show a radiator.

The amount of heat given out by this radiator will not be increased by increasing:

As the water flows through the radiator it loses heat and cools down so the water leaving the radiator is colder than that entering. The heat lost is related to the temperature of the radiator, increasing rate of flow will increase the average temp of the radiator.

With a higher specific heat capacity the temp drop of the liquid will be less so the average temperature of the radiator is increased.


Total Score:

All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. Any unauthorised copying or posting of materials on other websites is an infringement of our copyright and could result in your account being blocked and legal action being taken against you.