Thermal Physics Tutorial 1 - Heat Flow


You will be familiar with conduction, convection, and radiation from GCSE.  You did it in Year 10 (and you probably weren't listening?).  If you want to revise the topic, click HERE.


Conduction involves the flow of heat energy in a solid material.  As well as the vibration of molecules, electrons are involved.  This explains why metals are good conductors of heat.  Materials that do not conduct electricity do not conduct heat well.  Liquids and gases can conduct heat, but are very poor at doing so.


Models of conduction are not easy and are well beyond the scope of these notes. 



This occurs in fluids (liquids and gases).  It cannot happen in solid materials.  The particles in a fluid near a heat source vibrate more strongly.  Therefore they move apart, to occupy more space.  Therefore the density decreases, and the fluid rises.  As the particles rise higher, the heat is transferred to cooler particles.  Therefore the space occupied by the molecules decreases and the density rises.  The denser fluid falls.



The particles themselves do not change density, just the fluid which consists of the particles.



All hot bodies radiate heat energy in the form of electromagnetic radiation.  Thermal (infra-red) radiation behaves just like light:

Infra-red radiation has a wavelength of above 650 nm.


Shiny surfaces reflect infra-red radiation.  Matt-black surfaces emit and absorb radiation particularly well.


Heat pipes are often found in computers to transfer the heat generated by the processors to the cooling fan.  Here is a picture of some heat pipes in a laptop computer.



They use all three processes of heat transfer.  If you want to find out more, click HERE.


Heat Flow

Heat is the transfer of energy, a flow of energy from a hot body to a cooler body.  Heat is conducted by the movement of electrons as well as the vibration of moleculesGood conductors of electricity are good conductors of heat.  Heat must not be confused with internal energy, caused by the vibrations of atoms or molecules within a body. 


Temperature is a representation of internal energy, not heat flow.  The more internal energy there is, the higher the temperature.


At the start, the rod will be hot at one end and cold at the other end.  If we measure the temperature at the other end with a data-logger, we will see a graph like this:



When the heat flow into the rod is balanced by the heat flow out of the rod, we say that the system is in steady state.  The temperature in the rod varies in a linear way along the length of the rod.


Heat flows can be modelled in the same way as electric circuits:


Question 1

Professor Turner warned his first year university class, “…it is internal energy.  If I catch any of you calling it ‘heat’, I will personally come out and thump you. 

Can you explain the difference so that none of his students will get into trouble with the professor?



We can increase the internal energy by:

Some houses have electric central heating using night storage heaters.  These consist of a large metal box full of bricks, in which there are electrical elements.  The elements are switched on at night.  Electricity is cheaper at night, as power stations running at half load run less efficiently.



The elements transfer electrical energy to heat.  The heat flows into the bricks, increasing their internal energy.  The temperature rises.  When the electricity is turned off in the morning, the internal energy in the bricks is used to heat the room. 

The rate of heat flow into the room can be controlled with flaps.


Brownian motion can be used as a model to show the vibration of molecules. 



If you look at smoke particles in a cell under the microscope, you can see the particles jiggling about randomly as they are bombarded by air molecules.


Heat capacity

Heat capacity is the quantity of heat required to raise the temperature of a unit mass through a unit temperature rise.  It is given the physics code c.   The formula associated with this is:


         Heat flow (J) = mass (kg) × specific heat capacity (J kg-1 K-1) × temperature change (K)




Question 2

 Water has a specific heat capacity of 4200 J kg-1 K-1.  What is the amount of energy that is needed to bring 1.5 kg water to the boil from 20 oC


Question 3

 Here are the results of an experiment:

Joulemeter reading at the start

31225 J

Joulemeter reading at the end

43120 J

Temperature at the start

23 oC

Temperature at the end.

58 oC

Mass of the material

1.25 kg

Time taken

250 s

(a) Which piece of data is irrelevant?

(b)  What is the specific heat capacity of the material? 




In the exam:

Watch out for conversions.  Make sure that your mass in the equation is in kilograms.  It does not matter whether the temperature is in Celsius or Kelvin.  The steps are the same, and it’s the difference that matters.


Missing these conversions is a very common bear trap


We can measure the specific heat capacity of a metal in a simple experiment as shown in the diagram below:



We measure the voltage and current, which gives us power.  We should keep the voltage and current steady.  We then measure the temperature and time.


We can easily work out the energy supplied by:


E = VIt


We then need to plot the graph of temperature change against energy:




From this graph we can work out the specific heat from the gradient.



Specific heat with two materials

You may well find yourself having to work with more than one material.  An example of this may be:

Although this may appear more complex, it's not that much harder than with a single material.  


Suppose we have mass m1 of a liquid of specific heat c1 in a calorimeter (metal beaker) of mass m2 and specific heat c2. We want to change the temperature by Dq. To do this we need to supply DQ joules of energy.



For the liquid:

For the calorimeter:


The total energy supplied is:


To reduce uncertainty, the calorimeter needs to be well insulated around its sides.  It also needs to have a lid to reduce the heat loss by convection.


Here is a table of specific heat capacities of some materials:



The picture shows some oil in a copper calorimeter (a metal beaker).   As the oil gets hot, so will the calorimeter, so some energy will go to the oil, while some goes to the calorimeter.  We will use this for the worked example:


Worked Example

A copper calorimeter has a mass of 0.20 kg.  Oil, of mass 0.12 kg, is placed into the calorimeter.  The temperature of both the oil and the calorimeter is 20 oC.  15 kJ of energy is supplied to the oil and calorimeter, and the final temperature is 50 oC.  What is the specific heat capacity of the oil?  (c for copper = 381 J kg-1 K-1)



  DQ = mcDq

Calculate the energy taken by the calorimeter:

DQ = 0.20 kg × 381 J kg-1 K-1 × 30 K = 2286 J


The remaining energy is used to heat the oil.  So take 2286 J from 15000 J


Q = 15000 J - 2286 J = 12714 J


Now use this to calculate the specific heat capacity of the oil c' ("c-prime"):


c' =    Q       =    12714 J   

       mDq        0.12 kg × 30 K


c' = 3531 = 3500 J kg-1 K-1 (data are to 2 significant figures, so 2 s.f. is appropriate) 


Now let us see what happens if we drop a hot lump of metal into some cool liquid.  The liquid will have a low starting temperature, while the metal has a high starting temperature.  The key to solving a problem like this is the energy flow DQ from the metal is the same as the energy flow into the liquid.  Let us look at that in the next example:


Worked Example

A lump of iron of mass 1.0 kg has a temperature of 500 K.  It is dropped into 10 kg water which has a temperature of 300 K.  Neglecting the heat capacity of the container, and assuming that the mass of water lost to steam is negligible, what is the final temperature of the iron and water?  (c for iron = 438 J kg-1 K-1 c for water = 4200 J kg-1 K-1.)

The temperature change is NOT 200 K.  The energy from the hot iron raises the temperature of the water.  The energy going out of the iron means its temperature will fall.


The same energy leaves the iron as enters the water (Conservation of energy).


DQ = 1.0 kg × 438 J kg-1 K-1  × -Dq1 = 10 kg × 4200 J kg-1 K-1 × Dq2


Now the end temperature is T.  It will be the same for the iron and the water.


For the iron:

-Dq1 = T - 500 K (it will be negative because it's a temperature drop);


For the water:

Dq2 = T - 300 K (it will be positive because it's a temperature rise)

Now substitute:

1.0 kg × 438 J kg-1 K-1× -(T - 500 K) = 10 kg × 4200 J kg-1 K-1 × (T - 300K)


438 J K-1 ×  (-T + 500 K) = 42000 J K-1 × (T - 300K)


-438 T + 219 000 J = 42000 T - 1.26 × 107 J


-42000 T  - 438 T = -1.26 × 107 J - 2.19 × 105 J


42438 T = 1.2819× 107 J


T = 302 K (to 3 significant figures).



In this case, it is important to use absolute temperatures.  You will notice that the change in the temperature of the water is not very much.  Water has a particularly high specific heat capacity, which means you need to put in a lot of heat to get even a small rise in temperature.  That is why water is good at cooling things.



Latent Heat

The specific latent heat is the energy to change the state of a unit mass of liquid without a temperature change.  There is a value for specific latent heat during:

Whichever of these latent heats we are using, the calculation is the same.  The code for latent heat is L. 


  Energy flow = mass × specific latent heat.


The units are Joules per kilogram (J kg-1).


For water, the specific latent heat of fusion, lm = 334 000 J/kg.  The specific latent heat of vaporisation is rather higher, lv = 2.3 x 106 J/kg


Question 4

(Harder) Calculate how much energy is needed to melt, bring to the boil, and boil away 0.5 kg water.  How long would this take a 2 kW kettle? 



A change of state graph shows the idea of latent heat.


The rise in temperature is a reflection of the increase in internal energy.  In simple terms, internal energy is about vibration of molecules; the greater the vibration of molecules, the higher the internal energy.  When materials change state, work has to be done to break bonds between molecules.  Also most materials expand, so work has to be done against external forces.  So the latent heat represents the sum of these two:



While the material is changing state, there no increase in internal energy, so the temperature stays the same.

The reverse is true.  As a gas condenses, heat is given out.  This heat can be made to do a job of work.  However, as the gas condenses, there is no change in temperature.  Nor is there a change in temperature as the liquid freezes to a solid.


While most work has been done on water (as it’s very common and convenient to use), the same models can be applied to other substances.  Investigating iron would not be easy, as it melts at over 1000 oC and boils at about 3000 oC.


Hot materials can do jobs of work, which is studied in detail in a discipline called Thermodynamics (see the option Applied Physics).  The steam engine below is an example of how hot fluids are used to do very large amounts of work.



You can also see that a considerable amount of heat is being lost as waste steam.