## Section22.1Specific 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}.\label{eq-heat-capacity-definition}\tag{22.1.1} \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, in Physics, we now use 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. (22.1.1) 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{22.1.2} \end{equation}

Amount of heat needed to change temperature depends on other properties as well. 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 $c_P\text{.}$

### Subsection22.1.1Molar Specific Heat

Sometimes specific heat is defined per mole, especially in Chemistry, rather than per unit mass. Then, we call that 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{22.1.3} \end{equation}

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

### Subsection22.1.2Process-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.

### Subsection22.1.3Quantifying 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*}

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: $c_\text{Cu} = 0.379\text{J/g.K}\text{.}$

Hint

Use $Mc\Delta T\text{.}$

$136,000\ \text{J}\text{.}$

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

Therefore

\begin{align*} Q = \amp = 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{align*}

### Subsection22.1.4(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*}

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 $9\text{K}\text{?}$

Hint

Use the integral definition.

$52\ \text{J}\text{.}$