A$\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{V}$emf automobile battery has a terminal voltage of$\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{V}$when being charged by a current of$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{A}$.(a) What is the battery’s internal resistance?(b) What power is dissipated inside the battery?(c) At what rate (in ºC/min) will its temperature increase if its mass is$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{kg}$and it has a specific heat of$\mathbf{0}\mathbf{.}\mathbf{300}\mathbf{kcal}\mathbf{/}{\mathbf{kgx}}^{\mathbf{0}}\mathbf{C}$, assuming no heat escapes?

In this problem we consider a car battery with emf of $\mathrm{E}=12\mathrm{V}$that has a terminal voltage of $\mathrm{V}=16\mathrm{V}$when it is charged by a current of $\mathrm{l}=10\mathrm{A}$.

(a)

The terminal voltage of the charged battery is given as

$\mathrm{V}=\mathrm{E}+\mathrm{lr}$

$$\begin{gathered} \mathrm{r}=\frac{\mathrm{V}-\mathrm{E}}{\mathrm{l}} \\ \mathrm{r}=\frac{\left(16\mathrm{V}-12\mathrm{V}\right)}{10\mathrm{A}} \\ \mathrm{r}=0.4\mathrm{\Omega } \end{gathered}$$

Therefore, battery’s internal resistance is $\mathrm{r}=0.4\mathrm{\Omega }$.

(b)

The power dissipated inside the battery is the rate of thermal energy loss on the internal resistance, which is

$$\begin{gathered} \mathrm{P}={\mathrm{l}}^2\mathrm{r} \\ \mathrm{P}={\left(10\mathrm{A}\right)}^2\times \left(0.4\mathrm{\Omega }\right) \\ \mathrm{P}=40\mathrm{W} \end{gathered}$$

Therefore, power dissipated inside the battery is $\mathrm{P}=40\mathrm{W}$.

(c)

The rate of temperature increase is obtained from the power because power in this case is the rate of change in heat and heat is calculated as

$∆\mathrm{Q}=\mathrm{mc}∆\mathrm{T}$

where $\mathrm{m}=20\mathrm{kg}$is the mass of the resistance, $\mathrm{c}=0.3\mathrm{kcal}/\mathrm{kg}$.$°\mathrm{C}$is the specific heat. Specific heat can be written in terms of joules, using, $1\mathrm{kcal}=4184\mathrm{J}$as

$$\begin{gathered} \mathrm{c}=0.3\mathrm{kcal}/\mathrm{kg}.°\mathrm{C} \\ \mathrm{c}=0.3\times \left(4184\mathrm{J}\right)/\mathrm{kg}.°\mathrm{C} \\ \mathrm{c}=1255.2\mathrm{J}/\mathrm{kg}.°\mathrm{C} \end{gathered}$$

The change in temperature is then

role="math" localid="1655958286620" $$\begin{gathered} P=\frac{∆\mathrm{Q}}{∆\mathrm{t}} \\ \mathrm{P}=\mathrm{mc}\frac{∆\mathrm{T}}{\mathrm{t}} \\ \frac{∆\mathrm{T}}{\mathrm{t}}=\frac{\mathrm{P}}{\mathrm{mc}} \\ \frac{∆\mathrm{T}}{\mathrm{t}}=\frac{40\mathrm{W}}{\left(20\mathrm{kg}\right)\times \left(1255.2\mathrm{J}/\mathrm{kg}.°\mathrm{C}\right)} \\ \frac{∆\mathrm{T}}{\mathrm{t}}=1.59\times {10}^{-3}°\mathrm{C}/\mathrm{s} \\ \frac{∆\mathrm{T}}{\mathrm{t}}=0.096°\mathrm{C}/\min \end{gathered}$$

Therefore, rate of temperature change is $0.96°\mathrm{C}/\min$.