Chapter 5, Problem 18PEB

How many seconds will elapse between seeing lightning and hearing the thunder if the lightning strikes 1 mi (5,280 ft) away and the air temperature is 90.0°F?

To determine

The time difference between seeing lightning and hearing the thunder if the air temperature is $90.0°\text{ F}$ and the distance between the person and point of lightning is $1\text{ mi}$ or $5280\text{ ft}$.

Answer to Problem 18PEB

Solution:

$4.59\text{ s}$

Explanation of Solution

Given data:

The air temperature is $90.0°\text{ F}$ and the distance between the person and the point of lightning is $5280\text{ ft}$.

Formula used:

The expression to convert temperature in Fahrenheit to Celsius,

$T\left(°\text{ C}\right)=\left(°\text{ F}-32\right)\frac{5}{9}$

Write the expression for the velocity of a sound wave at the present temperature.

$v_{T\left(P\right)}=v_0+\left(\frac{2.00}{°\text{ C}}\right)\left(T_P\right)$

Here, $v_0$ represents the velocity of the sound wave at $0°\text{ C}$, and $T_P$ represents the present temperature.

Write the expression for the time.

$t=\frac{d}{v}$

Here, $d$ and $v$ are the distance and velocity respectively.

Explanation:

When the event occurs, it emits light and sound waves. The sound wave and light wave travelsby the speed of sound and light in air respectively. Take the case of travel of sound from the origin point to the person. The temperature of the air is $90.0°\text{ F}$. Convert it into degree Celsius.

$T\left(°\text{ C}\right)=\left(°\text{ F}-32\right)\frac{5}{9}$

Substitute $90.0°\text{ F}$ for $°\text{ F}$.

$\begin{array}{c}T\left(°\text{ C}\right)=\left(90.0-32\right)\frac{5}{9}\\ =32.2°\text{ C}\end{array}$

The sound is travelling at an air temperature of $32.2°\text{ C}$. Calculate the velocity of sound at this temperature.

$v_{32.2°\text{ C}}=v_0+\left(\frac{2.00\text{ ft/s}}{°\text{ C}}\right)\left(T_P\right)$

Here, $v_{32.2°\text{ C}}$ is the speed of sound at an air temperature of $32.2°\text{ C}$.

Substitute $1087\text{ ft/s}$ for $v_0$, and $32.2°\text{ C}$ for $T_P$.

$\begin{array}{c}{v}_{32.2°\text{ C}}=\left(1087\text{ ft/s}\right)+\left(\frac{2.00\text{ ft/s}}{°\text{ C}}\right)\left(32.2°\text{ C}\right)\\ =1087\text{ ft/s}+64.4\text{ ft/s}\\ \approx 1151\text{ ft/s}\end{array}$

The sound of the thunder travels at a speed of $1151\text{ ft/s}$. Calculate the time taken by sound to reach the person standing at a distance of $5280\text{ ft}$ from the point of origin of sound.

$t_{\text{sound}}=\frac{d}{v_{32.2°\text{ C}}}$

Substitute $5280\text{ ft}$ for $d$, and $1151\text{ ft/s}$ for $v_{32.2°\text{ C}}$.

$\begin{array}{c}{t}_{\text{sound}}=\frac{5280\text{ ft}}{1151\text{ ft/s}}\\ =4.59\text{ s}\end{array}$

Now, consider the case of travel of light from the source to the person. The velocity of light in air is $9.836\times {10}^8\text{ ft/s}$. Calculate the time taken by the light to reach the person.

$t_{\text{light}}=\frac{d}{v_{\text{light}}}$

Substitute $5280\text{ ft}$ for $d$, and $9.836\times {10}^8\text{ ft/s}$ for $v_{\text{light}}$.

$\begin{array}{c}{t}_{\text{light}}=\frac{5280\text{ ft}}{9.836\times {10}^8\text{ ft/s}}\\ =5.37\times {10}^{-6}\text{ s}\end{array}$

Thus, calculate the elapsed between seeing and hearing the event.

$t=t_{\text{sound}}-t_{\text{light}}$

Substitute $4.59\text{ s}$ for $t_{\text{sound}}$, and $5.37\times {10}^{-6}\text{ s}$ for $t_{\text{light}}$.

$\begin{array}{c}t=4.59\text{ s}-5.37\times {10}^{-6}\text{ s}\\ \approx 4.59\text{ s}\end{array}$

Conclusion:

The time elapsed between seeing lightning and hearing the thunder is $4.59\text{ s}$.