Question

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?

Short answer

a) Internal resistance of battery is$\mathrm{r}=0.4\mathrm{\Omega }$.

b) Power dissipated inside the battery is$\mathrm{P}=40\mathrm{W}$.

c) Rate of change of temperature is $0.096°\mathrm{C}/\min$.

Step-by-step solution

Calculation for battery’s internal resistance.

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 $\mathrm{r}=0.4\mathrm{\Omega }$.

Calculation for the power dissipated inside the battery.

(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 $\mathrm{P}=40\mathrm{W}$.

Calculation for the rate of change of temperature change.

(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 $0.96°\mathrm{C}/\min$.