Does the internal resistance of a battery dissipate heat as expected? A common real battery is often modelled (to a good approximation) as a perfect voltage source having emf $\varepsilon$ with a series internal resistance $r$. So, when a current $I$ is drained from it, the voltage across the terminals is $\varepsilon - I r$.
Now, does it follow that necessarily the power dissipated as heat equals $I^2 r$? If so, is the internal resistance due to some actual component within a battery?
(This would certainly be the case if the chemical processes inside the battery actually generated an emf $\varepsilon$ and something inside actually worked as a resistance. However, it need not be true if for some reason the "actual" emf is actually current-dependent and so there wouldn't be an exact $I r$ voltage drop across any part of the battery as to dissipate $I^2r$ power.)
 A: Results for an experiment to measure the voltage across the terminals of a battery$v$ as a function of the current passing through the battery $i$ many look something like this.

Where $v_{oc}$ is the open circuit terminal voltage and $i_{sc}$ is the current when the terminals are short circuited.
This is a straight line graph of the form $v = v_{oc} - ki$
When the terminals are short circuited then $0=v_{oc} - k i_{sc} \Rightarrow k = \dfrac {v_{oc}}{i_{sc}}$
So the abstraction which is used is to say that the $v_{oc}$ is the emf $\mathcal{E}$ of the battery and as $k$ has the dimensions of a resistor make $k = \dfrac {v_{oc}}{i_{sc}} = r$ the internal resistance of the battery.
Note that it is minus the gradient of the battery's $v-i$ characteristic.  
The power dissipated in the battery is $i^2r$.
There is no actual resistor in the battery (unless your teacher or lecturer has put one into a black box) rather the internal resistance represents as a lumped component the battery's resistance to the flow of current.
It is not a constant for a battery and changes with the battery's age and the current passing through the battery.
It is a useful approximation to a real voltage source.
A: An argument I found is actually the energy conservation.
Let's model your idea with an idealized battery with the following features:

*

*internal resistance $r_0$

*current-dependent emf during discharging: $\varepsilon(I) = \varepsilon_0 - k I$

*charge capacity $Q$

*no drop of emf nor rise of internal resistance after a partial discharge (not as essential assumption but simplifying the calculations)

From the first two points it follows that the voltage across the terminals is $$U(I) = \varepsilon(I) - r_0 I = (\varepsilon_0 - kI) -r_0 I = \varepsilon_0 - r I.$$
Here $r = k+r_0$ is the apparent inner resistance but the heat produced inside the battery is actually given only by $r_0 I^2$. The emf has an apparently constant value $\varepsilon_0$.
Now connect the battery to a circuit with a resistor with the resistance $R$ until it totally discharges. Then, the total produced heat is
$$W = (r_0 + R) I^2 t = (r_0 + R) I Q = (r_0 + R) \frac{\varepsilon_0}{r + R} Q = \frac {r_0 + R}{k + r_0 + R} \,\varepsilon_0 Q.$$
Notice that the amount of this heat depends on what external resistor $R$ you used: it ranges from $W_\text{min}=\frac {r_0}{k + r_0} \,\varepsilon_0 Q$ for a curcuit shorted by an ideal wire up to  $W_\text{max}=\varepsilon_0 Q$ for $R\gg r$. However, the total produced heat should be independent of $R$ since it is given by the total chemical energy that was stored in the battery and which we wasted. This paradox does not happen if and only if $k=0$, i.e., if the emf does not decrease with the current.
