# Changing systems with heat

Read Chapter 4

### Heat Capacities

Heat capacity: (limiting ratio of) heat supplied, $\Delta Q$, divided by the change in temperature, $\Delta T$. $$C = \frac{\delta Q}{dT}.$$

- $\Delta Q$ is an
*extensive*quantity: Usually we'll look instead at the*specific*heat capacity per kilomole. - We haven't specified a path, but we know the $\delta Q$ is not an exact differential.

The two kinds of paths we'll consider are: $$c_v \equiv \frac{1}{n} \( \frac{\delta Q}{dT} \)_v = \( \frac{\delta q}{dT} \)_v \ ;\ c_P \equiv \( \frac{\delta q}{dT} \)_P.$$

Now, consider the internal energy function $u$:

- $u$
*is*a state function, - $\Rightarrow$ specifying any 2 thermodynamic parameters is enough to uniquely determine the state and any function of the state.
- Your choice! $u(P,v)$, or $u(T,v)$, or $u(T,P)$.

Choosing $u(T,v)$: $$du = \( \frac{\partial u}{\partial T} \)_v dT + \( \frac{\partial u}{\partial v} \)_T dv.$$

For a *constant volume* process, the second term vanishes, leaving:
$$\( du \)_v = \( \frac{\partial u}{\partial T} \)_v dT.$$

According to the 1st Law, $du=\delta q -\delta w$. For a reversible process $\delta w = P\,dv$: $$du = \delta q-P\,dv \Rightarrow \( du \)_v = \delta q.$$

Taking the last two expressions for $du$ in **constant-volume
processes**, gives
$$\delta q = \( \frac{\partial u}{\partial T} \)_v dT.$$

Or, $$\( \frac{\delta q}{\partial T} \)_v = \( \frac{\partial u}{\partial T} \)_v.$$

So,

$$c_v = \( \frac{\partial u}{\partial T} \)_v.$$

We shall shortly see that,** for an ideal gas**, the internal energy actually
depends only on temperature, i.e. $u(T,v) = u(T)$, not on volume at all. When
this is true, then we can axe the subscript from the partial derivative, and
also dispense with the partial derivatives, and just write:
$$c_v = \frac{du}{dT} \text{ (ideal gas)}.$$

We can integrate to find the overall energy in a relatively easy fashion: $$\int du = u-u_0 = \int_{T_0}^{T} c_v dT.$$

**Problem (Carter 4-6 a):** *Pretend Oxygen is an ideal gas with $c_v=(5/2)R$.
Suppose that the temperature of 2 kilomoles of O${}_2$ is raised from 27 C
to 227 C. What is the increase in internal energy?*

**Problem (Carter 4-1)**: *The specific heat capacity $c_v$ of solids at low temperature
is given by the Debye T${}^3$ law:*
$$c_v = A \( \frac{T}{\theta}\)^3.$$

- $A$ is a constant where $A=19.4 \times 10^5$ J kilomole${}^{-1}$ K${}^{-1}$,
- $\theta$ is called the Debye temperature. For NaCl $\theta = 320$K.

### Image credits

Anne Petersen