Chapter 5, Problem 25Q

(a)

To determine

The wavelength of maximum emission of the volcano at a temperature of $320°\text{C}$.

Answer to Problem 25Q

Solution:

$4.89\text{ μm}$

Explanation of Solution

Given data:

The temperature of the volcano named Pele on Jupiter’s moon, Io, is $320°\text{C}$.

Formula used:

The conversion of the unit of temperature from Celsius to kelvin is done as follows:

$T_{\text{K}}=T_{\text{°C}}+273$

Here, $T_{\text{K}}$ is the temperature in kelvin and $T_{\text{°C}}$ is the temperature in Celsius.

The expression for Wien’s law is:

${\lambda }_{\text{max}}=\frac{0.0029\text{ m}\cdot \text{K}}{T}$

Here, $T$ is the temperature of the object and ${\lambda }_{\text{max}}$ is the wavelength of maximum emission of the object.

Explanation:

The conversion of the unit of temperature from Celsius to kelvin is done as follows:

$T_{\text{K}}=T_{\text{°C}}+273$

Substitute $320°\text{C}$ for $T_{\text{°C}}$.

$\begin{array}{c}\text{K}=320°\text{C}+273\\ =593\text{ K}\end{array}$

The expression for Wien’s law is:

${\lambda }_{\text{max}}=\frac{0.0029\text{ m}\cdot \text{K}}{T}$

Substitute $593\text{ K}$ for $T$.

$\begin{array}{c}{\lambda }_{\text{max}}=\frac{0.0029\text{ m}\cdot \text{K}}{593\text{ K}}\\ =4.89\times {10}^{-6}\text{ m}\left(\frac{{10}^6\text{ μm}}{1\text{ m}}\right)\\ =4.89\text{ μm}\end{array}$

Conclusion:

The wavelength of maximum emission of the volcano is $4.89\text{ μm}$.

(b)

To determine

The energy emitted by a square meter of the surface of Pele compared to that by a square meter of the surface of the moon, Io, if the temperature of Io’s surface is $-150°\text{C}$.

Answer to Problem 25Q

Solution:

A square meter of the surface of Pele emits $540.3$ times more energy than a square meter of the surface of Io.

Explanation of Solution

Given data:

The average temperature of the surface of Io is $-150°\text{C}$.

Formula used:

The conversion of the unit of temperature from Celsius to kelvin is done as follows:

$T_{\text{K}}=T_{\text{°C}}+273$

Here, $T_{\text{K}}$ is the temperature in kelvin and $T_{\text{°C}}$ is the temperature in Celsius.

The expression for Stefan Boltzmann law is:

$F=\sigma T^4$

Here, $T$ is the temperature of the object, $F$ is the energy flux per square unit per second and $\sigma$ is the Stefan constant $\left(5.67\times {10}^{-8}{\text{ Wm}}^{-2}{\text{K}}^{-4}\right)$.

Explanation:

The conversion of the unit of temperature from Celsius to kelvin is done as follows:

$T_{\text{K}}=T_{\text{°C}}+273$

Substitute $-150°\text{C}$ for $T_{\text{°C}}$.

$\begin{array}{c}\text{K}=-150°\text{C}+273\\ =123\text{ K}\end{array}$

The expression for Stefan Boltzmann law is:

$F=\sigma T^4$

Here, the energy per square unit per second will be same as the flux.

Use this relation to find the ratio of energy per square unit per second.

$\frac{F_{\text{pele}}}{F_{\text{Io}}}=\frac{T_{\text{pele}}^4}{T_{\text{Io}}^4}$

Substitute $123\text{ K}$ for $T_{\text{Io}}$ and $593\text{ K}$ for $T_{\text{pele}}$.

$\begin{array}{c}\frac{F_{\text{pele}}}{F_{\text{Io}}}=\frac{{\left(593\text{ K}\right)}^4}{{\left(123\text{ K}\right)}^4}\\ =540.3\end{array}$

Conclusion:

Therefore, Pele has a higher surface temperature, and each square meter of its surface emits $540.3$ times more energy compared to a square meter of the surface of Io.