Chapter 21, Problem 30P

(a)

To determine

The efficiency of the battery as an energy storage device.

Answer to Problem 30P

The efficiency of the battery as an energy storage device is $\underset{\_}{0.530}$.

Explanation of Solution

Write the expression for the energy taken by the battery.

    $E_{\text{in}}=P_{\text{in}}\Delta t_{\text{in}}$        (I)

Here, $E_{\text{in}}$ is the energy taken by the battery, $P_{\text{in}}$ is the input power, $\Delta t_{\text{in}}$ is the input time interval.

Write the expression for the energy used.

    $E_{\text{out}}=P_{\text{out}}\Delta t_{\text{out}}$        (II)

Here, $E_{\text{out}}$ is the output energy, $P_{\text{out}}$ is the output power, $\Delta t_{\text{out}}$ is the output time interval.

Write the expression for power.

    $P=\left(\Delta V\right)I$        (III)

Here, $\Delta V$ is the voltage, $I$ is the current.

Use equation (III) in (II) to solve for $E_{\text{out}}$.

    $E_{\text{out}}={\left(\Delta V\right)}_{\text{out}}{I}_{\text{out}}{\left(\Delta t\right)}_{\text{out}}$        (IV)

Use equation (III) in (I) to solve for $E_{\text{in}}$.

    $E_{\text{in}}={\left(\Delta V\right)}_{\text{in}}{I}_{\text{in}}{\left(\Delta t\right)}_{\text{in}}$        (V)

Write the expression for the efficiency.

    $\text{efficiency}=\frac{E_{\text{out}}}{E_{\text{in}}}$        (VI)

Use $1\text{V}=1\text{J}/\text{C}$  and $1\text{A}=1\text{C}/\text{s}$ relations to solve the equations.

Conclusion:

Substitute $2.3\text{V}$ for $\Delta V_{\text{in}}$, $13.5\text{mA}$ for $I_{\text{in}}$, $4.2\text{h}$ for $\Delta t_{\text{in}}$ in equation (V) to find $E_{\text{in}}$.

    $\begin{array}{c}{E}_{\text{in}}=\left(2.3\text{V}\right)\left(13.5\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(4.2\text{h}\times \frac{3600\text{s}}{1\text{h}}\right)\\ =\left(2.3\text{J}/\text{C}\right)\left(13.5\text{m}\text{C}/\text{s}\times \frac{{10}^{-3}\text{C}/\text{s}}{1\text{C}/\text{s}}\right)\left(4.2\text{h}\times \frac{3600\text{s}}{1\text{h}}\right)\\ =469\text{J}\end{array}$

Substitute $1.6\text{V}$ for $\Delta V_{\text{out}}$, $18\text{mA}$ for $I_{\text{out}}$, $2.4\text{h}$ for $\Delta t_{\text{out}}$ in equation (IV) to find $E_{\text{out}}$.

    $\begin{array}{c}{E}_{\text{out}}=\left(1.6\text{V}\right)\left(18\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(2.4\text{h}\times \frac{3600\text{s}}{1\text{h}}\right)\\ =\left(1.6\text{J}/\text{C}\right)\left(18\text{m}\text{C}/\text{s}\times \frac{{10}^{-3}\text{C}/\text{s}}{1\text{C}/\text{s}}\right)\left(2.4\text{h}\times \frac{3600\text{s}}{1\text{h}}\right)\\ =249\text{J}\end{array}$

Substitute $249\text{J}$ for $E_{\text{out}}$, $469\text{J}$ for $E_{\text{in}}$ in equation (VI) to find efficiency.

    $\begin{array}{c}\text{efficiency}=\frac{249\text{J}}{469\text{J}}\\ =0.530\end{array}$

Therefore, the efficiency of the battery as an energy storage device is $\underset{\_}{0.530}$.

(b)

To determine

The internal energy produced in the battery during one charge-discharge cycle.

Answer to Problem 30P

The internal energy produced in the battery during one charge-discharge cycle is $\underset{\_}{221\text{J}}$.

Explanation of Solution

Write the expression for the $E_{\text{out}}$.

    $E_{\text{out}}=E_{\text{in}}+\Delta E_{\text{int}}$        (VII)

Here, $\Delta E_{\text{int}}$ is the internal energy produced in the battery during one charge-discharge cycle.

Use equation (VII) to solve for $\Delta E_{\text{int}}$.

    $\Delta E_{\text{int}}=E_{\text{out}}-E_{\text{in}}$        (VIII)

Conclusion:

Substitute $249\text{J}$ for $E_{\text{out}}$, $469\text{J}$ for $E_{\text{in}}$ in equation (VIII) to find $\Delta E_{\text{int}}$.

    $\begin{array}{c}\Delta E_{\text{int}}=469\text{J}-249\text{J}\\ \text{=221}\text{J}\end{array}$

Therefore, the internal energy produced in the battery during one charge-discharge cycle is $\underset{\_}{221\text{J}}$.

(c)

To determine

The temperature rise during the cycle of charging and discharging.

Answer to Problem 30P

The temperature rise during the cycle of charging and discharging is $\underset{\_}{15.1°\text{C}}$.

Explanation of Solution

Write the expression for the heat energy.

    $Q=mc\Delta T$        (IX)

Here, $Q$ is the heat energy, $m$ is the mass of the battery, $c$ is the specific heat, $\Delta T$ is the temperature rise.

Use equation (IX) to solve for $\Delta T$.

    $\Delta T=\frac{Q}{mc}$        (X)

The internal energy $\Delta E_{\text{int}}$ is used as the heat energy and equation (X) becomes,

    $\Delta T=\frac{\Delta E_{\text{int}}}{mc}$        (XI)

Conclusion:

Substitute $221\text{J}$ for $\Delta E_{\text{int}}$, $15.0\text{g}$ for $m$, $975\text{J}/\text{kg}\cdot °\text{C}$ for $c$ in equation (XI) to find $\Delta T$.

    $\begin{array}{c}\Delta T=\frac{221\text{J}}{\left(15.0\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(9975\text{J}/\text{kg}\cdot °\text{C}\right)}\\ =15.1°\text{C}\end{array}$

Therefore, the temperature rise during the cycle of charging and discharging is $\underset{\_}{15.1°\text{C}}$.