Specific Heat

Section 22.1 Specific Heat

The amount of heat that can change the temperature of a body by one degree is called the heat capacity of the body. Therefore, if heat \(Q\) raises the temperature by \(\Delta T\) then the heat capacity would be

\begin{equation} \text{Heat Capacity } = \dfrac{Q}{\Delta T}.\tag{22.1} \end{equation}

Heat capacity is a misleading term, since it seems to imply that heat may be contained in the system rather than having only a transitory nature. Although the term “heat capacity” is still in use in certain disciplines such as Chemistry, but in Physics we don’t use this term any more. Instead, physicists prefer a related concept called specific heat.

Specific heat of a material is defined as the heat capacity of a material per unit mass. Dividing heat capacity in Eq. (22.1) by mass \(M \) of the sample will give the specific heat. Specific heat is usually denoted by the small letter \(c\text{.}\)

\begin{equation} \text{Specific Heat, }c = \dfrac{1}{M}\left( \dfrac{Q}{\Delta T}\right).\tag{22.2} \end{equation}

Amount of heat needed to change temperature depends on other properties in addition to amount of material, the temperature, and the material itself. For instance, heat to raise the temperature of a gas is different if you keep the volume constant versus if you keep the volume constant. Therefore, we have specific heat at constant volume, denoted by \(c_V\) and specific heat at constant pressure, denoted by \(c_P\text{.}\)

Subsection 22.1.1 Molar Specific Heat

Sometimes specific heat is defined per mole, especially in Chemistry, rather than per unit mass. Then, it’s called molar specific heat and denote it by the capital letter \(C\text{.}\) Say, it takes \(Q\) to raise the temperature of \(n\) moles of a substance by temperature \(\Delta T\text{.}\) Then,

\begin{equation} \text{Molar Specific Heat, } C = \dfrac{1}{n}\left( \dfrac{Q}{\Delta T}\right).\tag{22.3} \end{equation}

The molar specific heats at constant volume is denoted by \(C_V\) and that at constant pressure by \(C_P\text{.}\)

Subsection 22.1.2 Process-Dependence of Specific Heat

The specific heat of a substance depends on the manner in which the temperature of the substance is changed. We say that the specific heat is a process-dependent property.

For instance, different amount of heat will be required to raise the temperature \(\text{by }1^{\circ}\text{C}\) if you keep the pressure fixed during the change than if you keep the volume fixed, or if you kept neither of them fixed. That is, the amount of heat you would need depends on how you raise the temperature.

Two processes of particular interest, especially when we deal with gases, are the constant-pressure process (the isobaric process) and the constant-volume process (the isochoric process). The specific heats at constant pressure and at constant volume are denoted by \(c_P\) and \(c_V\) respectively.

\begin{align*} c_P \amp = \frac{1}{M}\left( \frac{Q}{\Delta T}\right)_P,\\ c_V \amp = \frac{1}{M}\left( \frac{Q}{\Delta T}\right)_V. \end{align*}

where subscript \(P\) and \(V\) are attached on the right side to emphasize the quantity that must remain constant during the process. The dimension of specific heat is Energy/(mass \(\times\) temperature) and therefore it has units of \(\text{J/kg.C or J/kg.K}\) in the SI system. Another commonly used unit is \(\text{cal/g.C}\text{.}\) Table in Engineering Toolbox Website

lists specific heats of some common substances.

Subsection 22.1.3 Quantifying Heat

Measurment of temperature along with specific heats table developed by experiments help us quantify heat involved in thermodynamic processes. We also note that for solids and liquids, \(c_P \approx c_V\text{,}\) so we do not make any difference when we work with solids. Only for gas, we will treat \(c_P \) and \(c_V\) differently.

Case: Specific Heat Independent of Temperature

For many substances, under normal conditions, specific heat is nearly independent of temperature over the temperature range under study. Therefore, you can deduce a formula for heat \(Q\) needed when a substance of mass \(M\) is heated so that its temperature rises from \(T_1\) to \(T_2\text{.}\)

\begin{align*} \amp Q = M c_P \left(T_2-T_1 \right) \ \ \text{(constant }p \text{ process)}\\ \amp Q = M c_V \left(T_2-T_1 \right) \ \ \text{(constant }V \text{ process)} \end{align*}

Case: Specific Heat Function of Temperature - Area Under the Curve

Often specific heat is a function of temperature, \(c(T)\text{.}\) In that case, to obtain heat needed for a finite change in temperature, we plot \(M\,c\) versus \(T\) and compute the area under the curve.

\begin{align*} \amp Q = \text{Area under }M\,c_P \text{ vs }T \ \ \text{(constant }p \text{ process)}\\ \amp Q = \text{Area under }M\,c_V \text{ vs }T \ \ \text{(constant }V \text{ process)} \end{align*}

Example 22.1. Heating a Block of Copper.

Find heat needed to raise the temperature of a \(2\text{-kg}\) block of copper from \(20^{\circ}\text{C}\) to \(200^{\circ}\text{C}\) at 1 atmosphere of pressure.

Data: \(c_\text{Cu} = 0.379\text{J/g.K}\text{.}\)

Answer.

\(136,000\ \text{J}\text{.}\)

Solution.

First, we note that for a copper block, \(c_P \) and \(c_V \) would not be very different. So we use just the symbol \(c \text{.}\)

Since, the specific heat of copper in the temperature range given in this problem is independent of temperature, we can simply use

\begin{equation*} Q = M\, c\, \left(T_2-T_1 \right). \end{equation*}

Therefore

\begin{equation*} Q = 2 \text{ kg}\times\left( 0.379\frac{\text{J}}{\text{g.K}}\times\frac{1000\text{ g}}{1 \text{ kg}}\right) \times 180\text{K} = 136,000\ \text{J}. \end{equation*}

Subsection 22.1.4 (Calculus) Quantifying Heat

More generally, the area under the curve described above is simply the computation of integration. Thus, we can write quantifying \(Q \) as integrals.

\begin{align*} \amp Q = M\,\int_{T_1}^{T_2}\,c(T)\,dT, \end{align*}

where \(c\) is the specific heat, which may, in general be a function of temperature, \(T\text{.}\) The specific heat at any temperature can similarly, be a derivative.

\begin{gather*} c = \dfrac{1}{M}\left( \dfrac{\partial Q}{\partial T}\right). \end{gather*}

Later on, when we discuss entropy, \(S\text{,}\) we will find that for a reversible process, we can write

\begin{equation*} dQ = T d S, \end{equation*}

which will give us another definition of specific heat as

\begin{equation*} c = \dfrac{T}{M}\left( \dfrac{\partial S}{\partial T}\right). \end{equation*}

Example 22.2. (Calculus) Low Temperature Study of Aluminum.

The specific heat of aluminum at constant pressure is found to vary with temperature as \(c_P(T) = \left(7.0\times 10^{-4} \text{J/g.K}^4 \right)T^3 \text{,}\) where \(T\) is the temperature expressed in the kelvin scale. How much heat is needed to raise the temperature of 50 grams of aluminum from \(5\text{K}\) to \(9\text{K}\text{?}\)

Answer.

\(52\ \text{J}\text{.}\)

Solution.

Note that the specific heat is temperature dependent. Therefore, we would have to integrate the specific heat to find heat required.

\begin{align*} Q \amp = m\int_{T_1}^{T_2} c_P(T) dT \\ \amp = 50\text{g}\times \left(7.0\times 10^{-4} \text{J/g.K}^4 \right)\int_{T_1}^{T_2} T^3 dT\\ \amp = \frac{3.5\times 10^{-2}}{4}(\text{J/K}^4 )\left(T_2^4 - T_1^4 \right) = 52\ \text{J}. \end{align*}

Exercises 22.1.5 Exercises

1. Low Temperature Specific Heat Varying with Temperature.

At very low temperatures the molar specific heats of metals vary as \(T^3\text{.}\)

\begin{equation*} \frac{C_P}{3R} = 77.9\left(\frac{T}{\Theta} \right)^3, \end{equation*}

where \(\Theta\) is called the Debye temperature. Here \(R\) is the universal gas constant and \(C_P\) is the molar specific heat for constant pressure. The Debye temperature of aluminum is \(420\)K. How much heat will be involved when the temperature of a \(3\)-kg aluminum block is raised from \(4\)K to \(20\)K?

Answer.

\(0.116\) J.

Solution.

Since \(C_P\) depends on the temperature \(T\) we integrate to obtain the heat exchanged. The calculation uses the value of \(R\) which is simpler in moles. Therefore, we will use moles for the quantity. There \(27\) g in each mole of Aluminum. Therefore the number of moles \(n\) in the given sample is

\begin{equation*} n = \frac{3000\ \textrm{g}}{27\ \textrm{g/mol}} = 111.11\ \textrm{mol}. \end{equation*}

\begin{align*} \Delta Q \amp = \int n C_P dT = 3\times (77.9/420^3)\times111.11 \times 8.31\int_{4}^{20} T^3 dT\\ \amp = 116\ \textrm{J}. \end{align*}

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