Chapter 17, Problem 50P

(a)

To determine

The resistance of wire at $273\text{K}$

Answer to Problem 50P

The resistance of wire at $273\text{K}$ is $2.675\times {10}^{-3}\Omega$

Explanation of Solution

Given Info: The tungsten wire has length $15.0\text{cm}$ and radius is $1.00\text{mm}$

Explanation:

Formula to calculate the resistance of wire A is,

$R_0=\frac{\rho L}{A}$

• • $R$ is the resistance of tungsten wire,
  • $\rho$ is the resistivity of tungsten
  • $L$ is the length of the tungsten wire,
  • $A$ is the area of circular cross section of the wire,

Formula to find the area is ,

$A=\pi r^2$

• • $r$ is the radius of circular cross section of the wire,

Substitute the above relation in the previous equation to rewrite R,

$R_0=\frac{\rho L}{\pi r^2}$

Substitute $5.6\times {10}^{-8}\text{Ω}\cdot \text{m}$ for $\rho$, $15.0\text{cm}$ for $L$, $3.14$ for $\pi$ and $1.00\text{mm}$ for $r$ in the above equation to find $R$

$\begin{array}{c}{R}_0=\frac{\left(5.6\times {10}^{-8}\text{Ω}\cdot \text{m}\right)\left(15.0\text{cm}\right)\left(\frac{{\text{10}}^{\text{-2}}\text{m}}{\text{1}\text{cm}}\right)}{\left(3.14\right){\left(1.00\text{mm}\right)}^2\left(\frac{{\text{10}}^{\text{-6}}{\text{m}}^2}{\text{1}{\text{mm}}^2}\right)}\\ =2.675\times {10}^{-3}\Omega \end{array}$

Conclusion: Therefore, the resistance of wire at $273\text{K}$ is $2.675\times {10}^{-3}\Omega$

(b)

To determine

The temperature of the wire

Answer to Problem 50P

The temperature of the wire is $1445.68\text{K}$

Explanation of Solution

Given Info: The tungsten wire has length $15.0\text{cm}$ and radius is $1.00\text{mm}$ The tungsten wire reaches the equilibrium temperature when it emits $75.0\text{W}$ radiation, The tungsten wire has length $15.0\text{cm}$ and radius is $1.00\text{mm}$ The tungsten wire reaches the equilibrium temperature when it emits $75.0\text{W}$ radiation.

Explanation:

Formula to calculate the radiating area of tungsten wire is,

$A=2\pi rl$

• • $A$ is the total surface area of a cylindrical wire,
  • $r$ is the radius of circular cross section of wire,
  • $l$ is the length of the wire,

Substitute $1.00\text{mm}$ for $r$, $15.0\text{cm}$ for $l$ and $3.14$ for $\pi$ in the above equation to find $A$,

  $\begin{array}{c}A=2\left(3.14\right)\left(1.00\text{mm}\right)\left(\frac{{\text{10}}^{\text{-3}}\text{m}}{\text{1}\text{mm}}\right)\left(15.0\text{cm}\right)\left(\frac{{\text{10}}^{\text{-2}}\text{cm}}{\text{1}\text{cm}}\right)\\ =9.42\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

The radiating area of tungsten wire is $9.42\times {10}^{-4}{\text{m}}^{\text{2}}$

Stefan’s law gives the formula to calculate the temperature

$P=\sigma Ae{T}^4$

• • $P$ is the net power radiated,
  • $\sigma$ is the Stephan’s constant,
  • $e$ is the emissivity of tungsten,
  • $T$ is the temperature of radiating body,

Substitute $75.0\text{W}$ for $P$, $5.669\times {10}^{-8}\text{W}\cdot {\text{m}}^{\text{-2}}\cdot {\text{K}}^{\text{-4}}$ for $\sigma$, $9.42\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A$ and $0.320$ for $e$ in the above equation to find $T$

$\begin{array}{c}T={\left(\frac{75.0\text{W}}{\left(5.669\times {10}^{-8}\text{W}\cdot {\text{m}}^{\text{-2}}\cdot {\text{K}}^{\text{-4}}\right)\left(9.42\times {10}^{-4}{\text{m}}^{\text{2}}\right)\left(0.320\right)}\right)}^{1/4}\\ =1445.68\text{K}\end{array}$

Conclusion: Therefore, the temperature of the tungsten wire is $1445.68\text{K}$

(c)

To determine

The resistance of wire at temperature $1445.68\text{K}$

Answer to Problem 50P

The resistance of wire at temperature $1445.68\text{K}$ is $0.017\Omega$

Explanation of Solution

Given Info: The tungsten wire has length $15.0\text{cm}$ and radius is $1.00\text{mm}$

Explanation:

Formula to calculate the resistance of wire is,

$R=R_0\left(1+\alpha \left(T-T_0\right)\right)$

• • $R$ is the resistance at temperature $T$,
  • $R_0$ is the resistance at some reference temperature $T_0$,
  • $\alpha$ is the temperature coefficient of resistivity of tungsten,

Substitute $2.675\times {10}^{-3}\Omega$ for $R_0$, $4.5\times {10}^{-3}{\text{K}}^{\text{-1}}$ for $\alpha$, $1445.68\text{K}$ for $T$, $273\text{K}$ for $T_0$ in the above equation to find $R$

$\begin{array}{c}R=\left(2.675\times {10}^{-3}\Omega \right)\left(1+\left(4.5\times {10}^{-3}{\text{K}}^{\text{-1}}\right)\left(1445.68\text{K}-273\text{K}\right)\right)\\ =0.01679\Omega \\ \approx 0.017\Omega \end{array}$

Conclusion: Therefore, the resistance of wire at temperature $1445.68\text{K}$ is $0.017\Omega$

(d)

To determine

The voltage drop across the wire

Answer to Problem 50P

The voltage drop across the wire is $1.13\text{V}$

Explanation of Solution

Given Info: The tungsten wire has length $15.0\text{cm}$ and radius is $1.00\text{mm}$

Formula to calculate the power radiated is,

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

• $\Delta V$ is the voltage drop across the wire,

Rewrite the above equation to find $\Delta V$ ,

$\Delta V=\sqrt{PR}$

Substitute $75.0\text{W}$ for $P$ and $0.017\Omega$ for $R$ in the above equation to find $\Delta V$

$\begin{array}{c}\Delta V=\sqrt{\left(75.0\text{W}\right)\left(0.017\Omega \right)}\\ =1.13\text{V}\end{array}$

Conclusion: Therefore, voltage drop across the wire is $1.13\text{V}$.

(e)

To determine

The reason for inefficiency of tungsten bulb as a light source.

Answer to Problem 50P

The high heat energy loss makes the tungsten bulbs inefficient.

Explanation of Solution

Efficiency of a light source is determined by the ratio of output light energy to input electrical energy. Higher this ratio, higher will be the efficiency.

Tungsten bulbs produce light by incandescence. Incandescence is the phenomenon in which the visible light is produced by heating the filament on passing electric current to very high temperature. By the law of conservation of energy, total input energy equals the total output energy. So by the heat energy emission , the ratio of useful output light energy to input electrical energy become smaller which means, efficiency of light source gets decreased.

Conclusion: The high heat energy loss makes the tungsten bulbs inefficient.