7.4 Thermal Properties of Materials

In the above examples, we have mentioned several thermal properties such as thermal capacity and specific heat. Now we wish to define them more carefully.

Specific heat is defined as: \[\begin{equation} c = \frac{\Delta O}{\Delta T \, M} \tag{7.17} \end{equation}\] where

\(c\) is the specific heat (J kg^-1 °C^-1),
\(\Delta Q\) is the heat flux (J),
\(\Delta T\) is a 1° change temperature (°C), and
\(M\) is a 1 kilogram system (kg).

We cannot measure the amount of heat a body can hold, so in a sense the term specific heat is misleading, but it is commonly used. It is simply the amount of heat which must be added per unit mass of material for a \(1^{\circ}C\) rise in temperature. What actually is changing is the internal energy of the system. Table 7.3 (after Monteith 1973) shows values of specific heat for different materials. See Stevenson (1979a) or Zemansky and Van Ness (1966) for a more formal definition.

The heat capacity of the system can then be obtained by multiplying the mass of the system by specific heat: \[\begin{equation} C = Mc \tag{7.18} \end{equation}\]

For instance, if we consider \(0.12 kg\) of granite and the same weight of peat soil as two systems (see Table 7.3, the heat capacity of each is the specific heat c times the mass \(M\) or \(96 J kg^{-1} {^{\circ}C^{-1}}\) for granite and \(226J kg^{-1} °C^{-1}\) the peat. To calculate the energy necessary to raise each system \(10°C\), we can employ equation (7.18) in the form \[\Delta Q = \Delta T \, M \, c\] The answer is found to 960 J for the granite and 2,200 J for the peat.

Table 7.3. Thermal properties of natural materials.

Material	Density	Specific heat	Thermal conductivity	Thermal capacity	Thermal diffusivity
	\(\rho\)	\(c\)	\(k\)	\(C_V=\rho c\)	\(k = \frac{k}{\rho c}\)
	kg m^-3 \(\times\) 10³	J kg^-1 °C^-1 \(\times\) 10³	Wm^-1°C^-1	J m^-3 °C^-1 \(\times\) 10⁶	m²s^-1 \(\times\) 10^-6
Granite	2.6	0.8	4.61	2.08	2.22
Quartz	2.66	0.8	8.8	2.13	4.14
Clay minerals	2.65	0.9	2.92	2.39	1.22
Ice	0.9	2.1	2.3	1.89	1.22
Old snow	0.5	2.1	0.29	1.05	0.28
New snow	0.1	2.1	0.08	0.21	0.38
Wet sand	1.6	1.3	1.68	2.08	0.81
Dry sand	1.4	0.8	0.17	1.12	0.15
Wet marsh soil	0.9	3.4	0.84	3.06	0.27
Peat soil	0.3	1.8	0.06	0.54	0.11
Still water	1	4.18	0.63	4.18	0.15
Still air	0.001	1	0.02	0.001	20
Organic matter	1.3	1.92	0.25		1
Fur	0.98		0.33
Mean body	0.98 - 1.05	3.42
Fat		1.88	0.14 - 0.20
Plant leaf
Wood	0.6	1.3	0.15	0.78	0.19

An equally valid way of considering the heat capacity of a system is to define the thermal capacity. It is the heat added per degree change per unit volume and thus is equal to the specific heat times the density of the material of the system (see Table 7.3). \[\begin{equation} c_v = \frac{\Delta Q}{\Delta T \, V} \tag{7.19} \end{equation}\] where

\(c_v\) is thermal capacity (\(J m^{-3}\)),
\(\Delta Q\) is the heat change in the system (\(J\)),
\(\Delta T\) is change in temperature (\(^{\circ}C\)), and
\(V\) is unit volume (\(m^3\)).

Thermal conductivity along with the notions of conductance and thermal resistance have already been discussed and defined in the section on conduction. The last parameter we wish to define is the thermal diffusivity \(K\) which is the ratio of the thermal conductivity \(K\) divided by density \(\rho\) times the specific heat.

\[\begin{equation} K = \frac{k}{\rho c} \bigg(\frac{J m^{-1} {^{\circ}C^{-1}} s^{-1}}{kg \, m^{-3} \, kg^{-1} {^{\circ}C^{-1}}}\bigg) \\ = m^2 s^{-1} \tag{7.20} \end{equation}\] The thermal diffusity is important in non-steady state conduction problems where thermal energy can be stored or transported as shown in the first example of heat flow in the soil.