Physics 6 Tutorial 14  Thermal Conduction
This tutorial is for students of the Welsh Board and Eduqas
Contents 
You will remember that metals are good thermal conductors, while nonmetals, liquids, and gases tend to be rather poor conductors (or good insulators). If you have forgotten it (or were offtask in that lesson), you can revise it HERE.
Conduction in any material can be explained using a model of molecules (represented by the balls in the picture below) joined with other molecules that are connected with bonds that act like springs.
We used a similar model to explain how wires obeyed Hooke's Law for tensile forces.
If we apply heat, the molecules vibrate with bigger vibrations and set their neighbours vibrating with bigger vibrations. These pass on the vibrations to their neighbours in turn. The bigger the vibrations, the hotter the material. If the vibrations are passed on easily, the material is a good conductor.
However this model does not explain why metals have a much better conductivity than non metals. The answer lies in the fact that metals can be modelled as being a lattice of ions in a sea of free electrons that can move easily between the ions.
As the metal ions are heated, the vibrations gain a bigger amplitude. Therefore there is a greater chance that the freemoving electrons will collide with the vibrating ions. The electrons then move off to other parts of the lattice and collide with other ions. The electrons transfer energy to the ions and these will vibrate with a larger amplitude. Thus the heat energy flows more rapidly from the hot part of the metal to the cool part of the metal.
We will look at how we can identify the factors that are involved in thermal conduction, and how they relate to each other. Consider a block of material that is mounted all the way around its edges with a perfect insulator (shown in the side view).
The only way that heat can pass is through the largest face that has an area A. The block has a thickness of Dx. The heat flow is constant, and the temperatures are at constant values (called steady state). The heat flow is from hot to cold, since heat never flows from cold to hot. The block looks like this from the front (not showing the perfect insulator).
Heat flow is defined as the amount of energy that passes in unit time. It is given the physics code:
The units are Joules per second (J s^{1}) or Watt (W).
Since the hot area is a distance x from the cold area, there is a temperature gradient:
The temperature gradient is given the physics code:
And the units are Kelvin per metre (K m^{1}). Temperature gradient is related to the temperatures by:
So we can say that the heat flow depends on two factors:
Area of the block;
Temperature gradient.
And we can write this as:
We then add in a constant of proportionality, l ("lambda", a Greek lower case letter 'l'):
The term l is the physics code for thermal conductivity, which is a property of the material. Thermal conductivity is defined as the heat flow per unit length per unit temperature. The units for thermal conductivity are W m^{1} K^{1}.
The thermal conductivity has a negative sign because the thermal gradient is negative. The heat flow is positive, so the extra negative sign cancels out the negative thermal gradient.
Show that the units for thermal conductivity are W m^{1} K^{1}. 
In the syllabus this equation is written slightly differently:
Note that the minus sign is ignored, as the temperature gradient is expressed as simply the temperature difference per unit length, without reference to the flow of heat from hot to cold. Because it's a temperature difference, you can use degrees Celsius, rather than Kelvin. Also the thermal conductivity is given the code K, rather than l.
Here are some typical thermal conductivities for different materials.
Material 
Thermal conductivity / W m^{1} K^{1} 
Copper 
385 
Aluminium 
238 
Steel 
60 
Concrete 
1.5 
Brick 
1.0 
Glass 
0.80 
Polythene 
0.50 
Wood 
0.30 
Air 
0.026 
Insulating Foam 
0.023 
A singleglazed window is made of glass 6.0 mm thick and is 1.3 m high and 2.0 m wide. The temperature in the room is 20 ^{o}C while the temperature outside is 5.0 ^{ o}C.
(a) Show that the temperature gradient is about 4200 K m^{1} (b) Looking up a suitable value on the table above, work out the heat flow through the window. 
This result is actually rather higher than what would be measured. This is because it assumes that the surface of the glass is actually at the same temperature as the room and the outside. This is not the case, as there is a thin layer of air on either side of the glass. Air is a rather poor conductor of heat, so the temperature gradient is rather less.
Architects and heat engineers do not use the thermal conductivity as shown above. Instead they use the Uvalue which is based on the measured heat flow out of a window or wall. The Uvalue is defined as:
the thermal conductivity per unit length.
We will explore the relationship between the thermal conductivity and the Uvalue
The Uvalue is measured as:
the rate of heat flow per unit area per unit temperature change.
In words, the relationship is:
In physics code:
This can be rearranged to give:
The units for the Uvalue are watts per square metre per Kelvin (W m^{2} K^{1})
Here are some Uvalues:
Building Elements 
UValue / W m^{2} K^{1} 
Solid Brick Wall 
2.0 
Cavity Wall (no insulation) 
1.5 
Insulated Wall 
0.18 
Single glazing 
4.8 to 5.8 
Double glazing 
1.2 to 3.7 
Triple Glazing 
1.0 
Solid timber door 
3.0 
UPVC 
2.2 
In some building regulation documents, the Rvalue is used. The Rvalue is simply the reciprocal of the Uvalue, and the units are K m^{2} W^{1}.
A singleglazed window is made of glass 6.0 mm thick and is 1.3 m high and 2.0 m wide. The temperature in the room is 20 ^{o}C while the temperature outside is 5.0 ^{ o}C.
(a) The room at is at a steady temperature. The temperature outside is 5.0^{ o}C.^{ }The heat flow through the window is measured at 300 W. If the Uvalue for the glass is 5.1 W m^{2} K^{1}, calculate the temperature in the room.
(b) Which data item is not relevant? 
The assumption made in the question above is that the window is mounted in a wall that is a perfect insulator, i.e. has a Uvalue of 0.
Electrical Analogy for Thermal Conductivity
We treat a building element like a window or a wall like a thermal conductor. The higher the thermal conductivity, the more heat flows through the element. It's like electricity. The higher the conductivity (or, strictly speaking, the conductance) of an electrical component, the greater the current for a given voltage. We can model thermal conductance like electricity flowing through resistors.
The table below shows the comparisons:
Thermal Conduction 
Electrical Conduction 
Heat flow (DQ/Dt) 
Current (I) 
Temperature Difference (Dq) 
Voltage (V) 
Thermal conductivity (k) 
Electrical conductivity (s) 
Distance of flow (Dx) 
Distance of flow (l) 
Relationship:

Relationship

The electrical quantities we are interested in are voltage and current. However we don't use resistance. Instead we use conductance. Conductance is the reciprocal of resistance.
G = R^{1}
It has the physics code G and the units are Siemens (S). In some textbooks you might see W^{1} or even "mho" ("ohm" written backwards). From this we can write familiar equations in terms of the conductance:
I = VG
G = I/V
V = I/G
The physical property conductivity is likewise the reciprocal to resistivity:
s = r^{1}
The units for conductivity are Siemens per metre (S m^{1}).
We know that the relationship for electrical resistivity is:
(a) What is the equivalent relationship for electrical conductivity? (b) Work out the equivalent relationship for thermal conductivity, K. (c) Draw a diagram to indicate the analogous terms between the two. 
When we combine series resistors, we know that:
R_{tot} = R_{1} + R_{2} + R_{3}...+ R_{n}
For parallel resistors:
R_{tot}^{1} = R_{1}^{1} + R_{2}^{1} + R_{3}^{1}...+ R_{n}^{1}
Now that we know that conductance is the reciprocal of resistance, we can write equations for series and parallel conductors.
For series conductors:
G_{tot}^{1} = G_{1}^{1} + G_{2}^{1} + G_{3}^{1}...+ G_{n}^{1}
For parallel conductors:
G_{tot} = G_{1} + G_{2} + G_{3}...+ G_{n}
The same can be done for thermal conductors.
If we have two building elements, for example a window set in a wall, heat can go through both the wall and the window. There are two paths. They are like the two parallel branches of a electrical circuit. Since we are using the analogy of conductance, we can say that the Uvalues add up:
Suppose, for the sake of simplicity, that the areas of the window and the walls were all the same. We use the analogy of parallel conductors:
The Uvalue of the insulated cavity wall is 0.18 W m^{2} K^{1}. The Uvalue of the single glazed window is 5.2 W m^{2} K^{1}. What is the total Uvalue? 
Rarely do we have a window and a wall panel that are exactly the same area. So we need to take into account the area that each element has. The product of the Uvalue and the area gives us the heat flow per unit temperature:
The heat flow per unit temperature has units of W K^{1}.
We can work out the Uvalue of a building element like a double glazed door by working out the area of the UPVC frame and the glass panels. Consider a door like this:
Worked Example The door in the diagram above consists of two double glazed panels of Uvalue 1.9 W m^{2} K^{1} of the dimensions shown. It is surrounded by UPVC of Uvalue 2.2 W m^{2} K^{1}. What is the total Uvalue of the door? 
Answer Work out the total area of the door: Area of the door = 1.8 m × 0.75 m = 1.35 m^{2}
Work out the areas of each glass pane: Area of top pane = 1.0 m × 0.57 m = 0.57 m^{2} Area of bottom pane = 0.70 m × 0.57 m = 0.40 m^{2} Total area of glass = 0.97 m^{2}
Now work out the area of the frame: Area of frame = 1.35 m^{2}  0.97 m^{2} = 0.38 m^{2}
Heat flow per unit temperature for the glass = 1.9 W m^{2} K^{1} × 0.97 m^{2} = 1.843 W K^{1}. Heat flow per unit temperature for the frame = 2.2 W m^{2} K^{1} × 0.38 m^{2} = 0.836 W K^{1}.
The heat flows add up, therefore: Total heat flow per unit temperature for the door = 1.843 W K^{1} + 0.836 W K^{1} = 2.679 W K^{1}
Uvalue of door = heat flow per unit temperature ÷ area = 2.679 W K^{1} ÷ 1.35 m^{2} = 1.98 W m^{2} K^{1} = 2.0 W m^{2} K^{1} (2 s.f. are appropriate here)

The temperature outside is 10^{ o}C while the temperature inside is 21 ^{o}C. What is the heat loss through the door in the example above? 
UValue and Thermal Conductivity
Earlier on in the tutorial, we defined the Uvalue as:
the thermal conductivity per unit thickness of the material.
We can also measure it as:
the rate of heat flow per unit area per unit temperature change.
We know that the heat flow is related to thermal conductivity with this relationship:
and heat flow is related to Uvalue by:
Therefore:
In Question 2, you looked at a singleglazed window that was 6.0 mm thick. The thermal conductivity was 0.80 W m^{1} K^{1}. What is the Uvalue? 
You can see that your answer to Question 7 is much higher than the Uvalue of a single glazed window, which was 5.1 W m^{2} K^{1}. This is because there is a layer of air either side of the window. Air is a poor conductor of heat. We will ignore convection. Consider our 6 mm pane of glass:
We have two layers of air, both of which have unknown thickness. However we can work out the Uvalue of the two layers of air. We will assume that the layers of air are the same thickness. We are also ignoring the effects of convection. We model the thermal conduction as current passing through series resistors. The idea is shown below:
For series resistors, the reciprocals of the conductance add up:
G_{tot}^{1} = G_{1}^{1} + G_{2}^{1} + G_{3}^{1}...+ G_{n}^{1}
Therefore:
U_{tot}^{1} = U_{air}^{1} + U_{glass}^{1} + U_{air}^{1}
So we can substitute:
5.1^{1} = 133^{1} + 2(U_{air}^{1})
and rearrange:
2(U_{air}^{1}) = 5.1^{1}  133^{1} = 0.196  0.00752 = 0.189
Turn it all upside down to get the final answer:
1/2U_{air} = 0.189^{1} = 5.30 W m^{2} K^{1}
Therefore each layer of air has a Uvalue of 10.6 W m^{2} K^{1}.
What is the thickness of the layer of air either side of the glass window? The thermal conductivity of air is 0.026 W m^{1} K^{1} 
It is entirely possible for house to be cooler inside than outside. I wrote these notes during the very hot July of 2018, and this was often the case that my house was several degrees cooler inside than it was outside.
Uvalues are an essential part of modern building regulations and there are several online tools for working out overall Uvalues for a whole house, taking into account not only walls, doors, and windows, but also the floors, ceilings and the roof. We want to keep the house warm at minimum cost because:
we want to have money to spend on other things (as the rent/mortgage allows);
heating houses leads to emission of carbon dioxide (our carbon footprint);
a warm house is welcoming;
a warm house requires less maintenance.
A cold house is not just unpleasant to be in. A cold house leads to condensation which leads to deterioration in the decoration of rooms. It encourages mould that not only damages the furniture and fitting, it also can cause health problems. When heating a house, we are paying to keep ourselves warm, not to provide a heated platform for jackdaws and feral pigeons to warm their bottoms on the roof.