We know from everyday experience that some things need more energy than others to raise their temperatures by the same amount. For instance, it takes more energy to heat water by one kelvin than the same mass of copper requires. The water molecules require more energy in order to move around fast enough to record a one kelvin rise than do the copper atoms, which mainly vibrate. The heat capacity, C, of an object is the amount of energy required to raise its temperature by one kelvin. The unit is kJ K^-1.

^{A boiling water. Image by Ken Boyd from Pixabay}

SPECIFIC HEAT CAPACITY, MOLAR HEAT CAPACITY AND LATENT HEAT

Heat capacity refers to a particular object of any mass. A more general quantity would be more useful. This is specific heat capacity, c, which refers to unit mass of a single substance. Each pure substance has its own particular value of specific heat capacity at a given temperature (though this value may change as the temperature of the substance varies). In heating a substance energy is given to its molecules, and unit masses of different substances have different numbers of molecules of different mass.

We can compare expressions for heat capacity C and specific heat capacity c. Let’s say we heat an object of mass m (kilograms) of a single substance by giving it energy ΔQ (kilojoules) so that the temperature rises by ΔT (kelvin or Celsius), then:

C = ΔQ/ΔT and c = ΔQ/mΔΤ

EXAMPLE

A 3.0 kW electric kettle is used to bring 1.50 kg of water to the boil from a starting temperature of 18.0 °C. Assuming all the energy goes into heating the water, calculate the amount of energy and the time required to boil the water. Specific heat capacity of water = 4.20 kJ kg^-1 K^-1.

ANSWER: Energy required = specific heat × mass × temperature rise

= 4.20 × 1.5 × 82 kJ

= 517 kJ

Time required = energy required/rate of heating

= 517 kJ/3.0 kJs^-¹

= 172 s, or 2 min 52 s

We can use another thermal quantity, the molar heat capacity, particularly for gases. As the name suggests, this is the energy required to heat one mole of the substance through one kelvin.

We have already seen that when a substance changes from one state to another, for example from solid to liquid, energy called latent heat required to produce the change. The latent heat does not cause any change in temperature. The ordering (arrangement) and the spacing of the molecules is changed but not the temperature. Also, most substances have different specific heat capacities depending on whether they are solid, liquid or gas. If we add energy to the substance at a constant rate, we can see how these differences in heat capacity affect the rate of heating, and we can identify the temperatures where changes of state occur. We assume no energy loss to the surroundings.

THERMAL CONDUCTIVITY

When one side of a body is heated, it takes time for the internal energy to spread (by thermal conduction) to the other side. That is, there is a time lag between the heated side of the body changing its temperature by a significant amount and a corresponding change occurring on the far side. Once thermal equilibrium has been reached, a temperature difference will be maintained through the object, assuming that heating continues and that energy can escape from the far side. The ease with which energy can transfer through the object depends on a property called thermal conductivity. However, as well as thermal conductivity, other factors determine how much energy passes through the object. We now consider all these factors.

Take the flow of energy through a block of material with a thickness of L metres. For energy to flow, there has to be a temperature difference between opposite faces of the block.

Assuming the block is of exactly the same material throughout, the temperature gradient at any point between the opposite faces will be the same. We have:

Temperature difference = temperature gradient/thickness of block

= T₁-T₂/L

The larger the temperature gradient, the greater the rate of flow of energy.

The flow will also depend on the area of cross-section A (m²) of the block. This is very similar to water passing through a pipe – the quantity of water flowing depends on the area of cross-section of the pipe and the pressure difference between the two ends. Transfer of thermal energy is also similar to flow of charge in an electrical conductor – electrical current is proportional to the cross-sectional area of the conductor and to the electric field (potential gradient) along the conductor.
The amount of energy that flows will depend on how well the material conducts it. Copper is a good conductor, while glass is a poor conductor. The rate of energy flow is proportional to the thermal conductivity, k, of the material. This dependence is the same as the way in which flow of electric charge depends on the electrical conductivity of a conductor. The rate of flow of energy is the quantity of thermal energy Q (kilojoules) transmitted, divided by time t (seconds). We have:

Q/t α T₁- T₂/L

Q/t α A

Q/t α k

So far, k has not been fixed, so we can malke it the constant of proportionality for the combined expression:

Q/t = kA (T₁- T₂)/L

Note that, in this equation, T₁ is greater than T₂ and that energy goes down a temperature gradient from the side of higher temperature to the side of lower temperature.

Units of k: k will have units which we work out from the equation. Q/t has units of energy/time, that is J s^-1 or W. A (T₁- T₂)/L has units of length² × temperature/length, that is m² K m^-1, equal to m K. (Note the space between m and K as otherwise the symbols would represent one thousandth of a kelvin.)

So: W = (units of k) × m K. So the units of k must be W divided by m K which equals W m^-1 K^-1.

^{Carnot engine diagram. Eric Gaba, Public Domain}

THERMAL FLOW OF ENERGY: CONDUCTION

FLOW ALONG A ROD: Suppose we heat one end of a long rod that is thermally lagged (i.e. surrounded with insulation). We assume that energy flow is the same at all points along the whole length. We can represent this flow by a series of parallel lines, just as we can illustrate the uniform flow of water through a pipe.

What is happening on the atomic scale? We have already said that we can think of the atoms as solid spheres connected in all directions by springs. When we heat one end of the rod, the atoms at that end vibrate with increasing amplitude. Energy from the vibrations of these atoms is passed on to neighbouring atoms. Eventually the amplitude of vibration of all the atoms in the solid increases.

After a time, the rod reaches a state of equilibrium, with just as much energy leaving the far end as enters the heated end. Energy is transferred thermally through the rod, which means that, as you would expect, there is a temperature gradient in the rod. The far end is cooler than the heated end. We can think of the difference in temperature as driving the energy through the rod, just as a difference in voltage drives an electric current through a wire.

Some of the energy in a metal is carried by free electrons. They move around faster at the hot end than at the cold end. By colliding with the lattice of ions (the spheres in our ball-and-spring model) they help rapid transfer of energy through the rod. We have already noted that copper, a typical metal, has a large thermal conductivity. This is because of its free electrons.

The vibrations of the atoms in the metal rod are transmitted as waves along the rod. However, just as sometimes light is described as the passage of particles called photons with no rest mass, so the passage of these vibrations can be also considered as the movement of particles with no rest mass, this time called ‘phonons’.

EXAMPLE

An iron bar 0.50 m long and a copper bar 1.2 m long are joined end to end. One end of the iron bar is kept at 80°C while the far end of the copper bar is maintained at 0°C by a mixture of ice and water. The outer surfaces of the bars are lagged so that there are no thermal energy losses. Both bars are of circular cross-section, diameter 0.16 m. At thermal equilibrium the temperature at the junction of the metals is T_j. Calculate T_j and the rate of energy flow. Thermal conductivity of iron = 75 W m^-1 K^-1, thermal conductivity of copper = 390 W m^-1 K^-1.

ANSWER: The rate of energy flow through each conductor is the same.

Rate of energy flow through the iron bar is:

Q/t = kA (T₁- T₂)/L

Q/t = k_iron A (T₈₀- T_j)/0.5 (watts)

= 75 × π × 0.08² × (80 – T_j)/0.5

= (3.02)(80 – T_j) W

The rate of energy flow through the copper bar is:

Q/t = kA (T₁- T₂)/L

Q/t = k_copper A (T_j- T₀)/1.2 (watts)

= 390 × π × 0.08² × (T_j– 0)/1.2

= (6.53)(T_j – 0) W

Equating gives:

(3.02)(80 – T_j) = 6.53(T_j- 0)

242 = 9.55T_j

T_j = 25.3 °C

Hence, Q/t = 165 Js^-1

Suppose that the rod in the example above is not lagged. In this case, energy is lost by thermal processes along the sides of the rod.

This is similar to water escaping from holes in the side of a hosepipe (as often used for watering gardens or orchards). The flow lines will no longer be parallel. More importantly, we can no longer assume a constant temperature gradient.

FLOW THROUGH A GLASS WINDOW

The thermal energy flow through glass windows is of great practical importance. Glass has a low thermal conductivity, comparable to that of brick. But because the thickness of glass in a window is so much than the thickness of a typical house brick, it is important to reduce the loss of energy through windows. The most common way is to use double glazing.

EXAMPLE

A room has a single glass window of length 2.2 m, height 1.2 m and thickness 5 mm. Assuming that the temperature in the room at the surface of the glass is 22 °C and that outside it is 3 °C, calculate the loss of energy from the room. Thermal conductivity of glass is 0.8 W m^-1 K^-1.

ANSWER: We use the equation:

Rate of energy flow = Q/t = kA (T₁- T₂)/L (watts)

= 0.8 × (2.2 × 1.2) × (22-3)/0.005

= 8026 W

(Note that it is not necessary to convert °C to K because the equation involves a temperature difference and one degree Celsius has the same magnitude as one kelvin.)

At this point we should question the model. We are losing over 8 kW of thermal energy from the room, and would need a very large heater to maintain the temperature at 22°C. Something must be wrong. Yet the numbers we have put in seem reasonable.

The error is that we have not allowed for layers of still air on either side of the window. The thermal conductivity of air is very low (approximately 0.025 W m^-1 K^-¹). Part of the temperature drop will be just inside the window and part will be just outside. Therefore the temperature difference between the opposite sides of the glass and the temperature gradient through the pane of glass will be rather less than we used in the calculation. The overall temperature profile is not as we have assumed.

Note that there has to be some temperature gradient at the two glass surfaces; otherwise, energy would be unable to enter or leave. We would need to take measurements to find the actual profile close to the window pane. Away from the windows of the room, energy is transferred mainly by convection.

Double glazing and thermal resistance

Most new windows around are now double glazed. The glass is commonly 4 mm thick, though the distance between the panes is less standard, perhaps 5 mm for windows or now, more usually, at least 10 mm for patio doors. Using the analogy of flow of charge through a conductor, we can see that our double-glazed unit can be modelled as three resistances in series: the resistances of glass, air and glass. So the thermal resistance R of double glazing will be the sum of the thermal resistances of the three layers. The thermal resistance R for one layer is defined as t/kA where t is the layer thịckness, A its area and k the thermal conductivity. With the three layers present, the summation is represented by:

∑t/kA

Where the symbol ∑ (capital Greek Ietter sigma), meaning ‘sum of’, refers to the three layers – glass, air and glass.

Now k_air = 0.025 W m^-¹ K^-¹ and k_glass = 0.8 W m^-1 K^-1. Therefore with windows of thickness 4 mm, separation 10 mm and area 1 m², we obtain:

Thermal resistance = (4 × 10^-³/0.8) + (10^-2/0.025) + (4 × 10^-3/0.8) K W^-1

= 0.41 K W^-1

For a single pane of glass, the thermal resistance will be 0.005 K W^-¹, but remember that there will be some convection in the air gap also.

^{EURO 68 wooden window profile with insulated glazing. Kozuch, CC BY-SA 3.}0

U-VALUES

U-values are quantities used by architects and heating engineers for working out thermal energy flows within buildings – in particular, energy losses through the windows and walls. The U-value of a particular thickness of material is its thermal transmittance, that is, the thermal energy flow through the material per unit area for a temperature difference of one degree. Hence it is given by:

U-value = rate of energy flow/(area × temperature difference)

Its units are W m^-2 K^-1. U-values are available for different types of walls, windows, floors and roofs, The values have been obtained by direct measurement rather than theoretical calculation. Thus thermal transmittance values are more realistic than thermal conductivity values as they take account of the actual composition of the building material and allow for any convection within hollow components.

The thermal transmittance from a room or building can be calculated by multiplying the U-value for each component of the surface structure by the corresponding area.

SUMMARY

After reading these chapters on ENERGY AND TEMPERATURE by starting with this, you should be able to:

Discuss the equivalence of work and energy and, in particular, the equivalence of work and the thermal energy required to heat a substance.
Describe the thermal energy transfer processes of conduction, convection and radiation.
State the advantages and disadvantages of different methods of measuring temperature.
Describe the difference between the thermodynamic (absolute or Kelvin) and Celsius scales of temperature.
Explain the terms: thermal capacity, specific heat capacity, thermal conductivity and U-value.
Carry out calculations on the change of temperature of objects using thermal capacity or specific heat capacity.
Calculate thermal resistance using t/kA for a single layer and ∑(t/kA) for a number of layers.
Use thermal conductivity (or U-value) for calculating thermal energy flow through materials.