Chapter 21, Problem 28P

(a)

To determine

The relationship between the diameter and the length of the wire.

Answer to Problem 28P

The relationship between the diameter and the length of the wire is $\underset{\_}{d^2=\left(4.77\times {10}^{-8}\right)l,\text{where }d\text{and}l\text{ are in meters}}$.

Explanation of Solution

Write the expression for the output power of the heater.

    $P=\frac{Q}{\Delta t}$        (I)

Here, $P$ is the power output, $Q$ is the heat energy stored, $\Delta t$ is the time.

Write the expression for the $Q$.

    $Q=mc\left(T_f-T_i\right)$        (II)

Here, $m$ is the mass, $c$ is the specific heat, $T_f$ is the final temperature, $T_i$ is the initial temperature.

Use equation (II) in (I) to solve for $P$.

    $P=\frac{mc\left(T_f-T_i\right)}{\Delta t}$        (III)

Write the expression for power relating the resistance.

    $P=\frac{{\left(\Delta V\right)}^2}{R}$        (IV)

Here, $\Delta V$ is the potential difference, $R$ is the resistance.

Use equation (IV) to solve for $R$.

    $R=\frac{{\left(\Delta V\right)}^2}{P}$        (V)

Write the expression for the resistivity.

    $\rho ={\rho }_0\left[1+\alpha \left(T_f-T_i\right)\right]$        (VI)

Here, $\rho$ is the resistivity, ${\rho }_0$ is the resistivity at reference temperature, $\alpha$ is the temperature coefficient.

Write the expression for resistance.

    $R=\frac{\rho l}{A}$        (VII)

Here, $l$ is the length of the wire, $A$ is the area of cross-section.

Write the expression for $A$.

    $A=\pi {\left(\frac{d}{2}\right)}^2$        (VIII)

Here, $d$ is the diameter of the wire.

Use equation (VIII) in (VII) to solve for $\frac{l}{d^2}$.

    $\begin{array}{c}R=\rho \frac{l}{\pi {\left(\frac{d}{2}\right)}^2}\\ \frac{l}{d^2}=\frac{\pi R}{4\rho }\\ d^2=\frac{4\rho }{\pi R}l\end{array}$        (IX)

Use the unit conversion $1\text{V}=1\text{J}/\text{C}$.

Conclusion:

Substitute $250\text{g}$ for $m$, $4186\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $100°\text{C}$ for $T_f$, $20°\text{C}$ for $T_i$, $4.00\text{min}$ for $\Delta t$ in equation (III) to find $P$.

    $\begin{array}{c}P=\frac{\left(250\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(4186\text{J}/\text{kg}\cdot °\text{C}\right)\left(100°\text{C}-20°\text{C}\right)}{\left(4.00\text{min}\times \frac{60\text{s}}{1\text{min}}\right)}\\ =349\text{J}/\text{s}\end{array}$

Substitute $349\text{J}/\text{s}$ for $P$, $120\text{V}$ for $\Delta V$ in equation (V) to find $R$.

    $\begin{array}{c}R=\frac{{\left(120\text{V}\right)}^2}{349\text{J}/\text{s}}\\ =\frac{{\left(120\text{J}/\text{C}\right)}^2}{349\text{J}/\text{s}}\\ =41.3\Omega \end{array}$

Substitute $1.50\times {10}^{-6}\Omega \cdot \text{m}$ for ${\rho }_0$, $0.4\times {10}^{-3}{\left(°\text{C}\right)}^{-1}$ for $\alpha$, $100°\text{C}$ for $T_f$, $20°\text{C}$ for $T_i$ in equation (VI) to find $\rho$.

    $\begin{array}{c}\rho =\left(1.50\times {10}^{-6}\Omega \cdot \text{m}\right)\left[1+\left(0.4\times {10}^{-3}{\left(°\text{C}\right)}^{-1}\right)\left(100°\text{C}-20°\text{C}\right)\right]\\ =1.55\times {10}^{-6}\Omega \cdot \text{m}\end{array}$

Substitute $1.55\times {10}^{-6}\Omega \cdot \text{m}$ for $\rho$, for $R$ in equation (IX) to find $\frac{l}{d^2}$.

    $\begin{array}{c}\frac{l}{d^2}=\frac{\pi \left(41.3\Omega \right)}{4\left(1.55\times {10}^{-6}\Omega \cdot \text{m}\right)}\\ =2.09\times {10}^7{\text{m}}^{-1}\\ d^2=\left(4.77\times {10}^{-8}\text{m}\right)l\end{array}$

Therefore, the relationship between the diameter and the length of the wire is $\underset{\_}{d^2=\left(4.77\times {10}^{-8}\right)l,\text{where }d\text{and}l\text{ are in meters}}$.

(b)

To determine

The volume of the nichrome wire is less than $0.500{\text{cm}}^3$ can be used for the heater

Answer to Problem 28P

$\underset{\_}{\text{Yes, for}V=0.500{\text{cm}}^3\text{of Nichrome wire with }l=3.65\text{m}\text{and}d=0.418\text{mm}}$ can be used for the water heater.

Explanation of Solution

Write the expression for the volume.

    $V=Al$        (X)

Here, $V$ is the volume.

Use equation (VIII) in (X) to solve for $V$.

    $V=\frac{\pi d^2}{4}l$        (XI)

Use the relation obtained from part (a),

    $d^2=\left(4.77\times {10}^{-8}\right)l$        (XII)

Use equation (XII) in (XI) to solve for $l$.

    $\begin{array}{c}l=\frac{4V}{\pi d^2}\\ l=\sqrt{\frac{4V}{\pi \left(4.77\times {10}^{-8}\text{m}\right)}}\end{array}$        (XIII)

Use equation (XIII) in (XII) to solve for $d$.

    $d={\left(\left(4.77\times {10}^{-8}\right)l\right)}^{1/2}$        (XIV)

Conclusion:

Substitute $0.500{\text{cm}}^3$ for $V$ in equation (XIII) to find $l$.

    $\begin{array}{c}l=\frac{4\left(0.500{\text{cm}}^3\times \frac{{10}^{-6}{\text{m}}^3}{1{\text{cm}}^3}\right)}{\pi \left(4.77\times {10}^{-8}\text{m}\right)}\\ =3.65\text{m}\end{array}$

Substitute $3.65\text{m}$ for $l$ in equation (XIV) to find $d$.

    $\begin{array}{c}d={\left[\left(4.77\times {10}^{-8}\text{m}\right)\left(3.65\text{m}\right)\right]}^{1/2}\\ =4.174\times {10}^{-4}\text{m}\\ \approx 0.418\text{mm}\end{array}$

Therefore, the volume $\underset{\_}{V=0.500{\text{cm}}^3\text{of Nichrome wire with }l=3.65\text{m}\text{and}d=0.418\text{mm}}$ can be used for the heater.