Question

A person yells across a canyon and hears an echo 2.5 seconds later. The air temperature is 21°C. How far away is the canyon wall?

Answer 1

Answer: The canyon wall is 850 m away from the person

Explanation:

Well, the speed of sound $V$ in air at $21 \° C$ is defined as $343.60 m/s$, this can be calculated by the following equation:

$V=\sqrt{\frac{\gamma R T}{M}}$ (1)

Where:

$\gamma=1.4$ is the Heat capacity ratio for air

$R=8.314 J/mol.K$ is the Universal Gas constant

$T=21 \° C+273.15=294.15 K$ is the temperature in Kelvin

$M=0.029 kg/mol$ is the air molar mass

$V=\sqrt{\frac{(1.4)(8.314 J/mol.K)(294.15 K)}{0.029 kg/mol}}$

$V=343.60 m/s$ (2)

Now that we know the speed of sound, we can use the following equation to find the distance between the person and the canyon wall:

$V=\frac{d}{t}$ (3)

Where:

$d$ is the distance between the person and the canyon wall

$t=2.5 s$ is the time it takes to the sound wave to travel from the person and then go back

Isolating $d$:

$d=V.t$ (4)

$d=(343.60 m/s)(2.5 s)$ (5)

Finally:

$d=850 m$

Answer 2

The distance at which the caynon wall is far away will be:

"850 m".

Heat and temperature

According to the question,

Heat capacity, $\gamma$ = 1.4

Universal gas constant, R = 8.314 J/mol.K

Temperature, T = 20°C + 273.15

                          = 294.15 K

Air molar mass, M = 0.029 kg/mol

Time, t = 2.5 s

We know the formula,

→ Speed of sound, V = $\sqrt{\frac{\gamma R T}{M} }$

By substituting the values,

                                   = $\sqrt{\frac{1.4\times 8.314\times 294.15}{0.029} }$

                                   = 343.60 m/s  

hence,

The distance be:

→ d = V.t

     = 343.60 × 2.5

     = 850 m

Thus the above response is correct.

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