Question

On a warm summer day \(\left( {31^\circ {\rm{C}}} \right)\), it takes 4.80 s for an echo to return from a cliff across a lake. On a winter day, it takes 5.20 s. What is the temperature on the winter day?

Short answer

The temperature on the winter day is \( - 13.11^\circ {\rm{C}}\).

Step-by-step solution

Understanding the speed of moving system

In this problem, the temperature on the winter day can be evaluated by using the relation of the speed of the moving system, which is equal to the ratio of total distance travel by the system to the time of travel.

Given data

The temperature on summer day is \({T_1} = 31^\circ {\rm{C}}\).

The time taken by echo to return from a cliff across a lake in summer days is \({t_1} = 4.80\;{\rm{s}}\).

The time taken by echo to return from a cliff across a lake in winter days is \({t_2} = 5.20\;{\rm{s}}\).

Evaluating the speed of sound during the summer day

The speed of sound during summer day is calculated below:

\({v_1} = 331 + 0.61\left( {{T_1}} \right)\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{v_1} = 331 + 0.61\left( {31^\circ {\rm{C}}} \right)\\{v_1} = 349.91\;{\rm{m/s}}\end{aligned}\)

The total distance travelled by an echo is calculated below:

\(d = {v_1}{t_1}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}d = \left( {349.91\;{\rm{m/s}}} \right)\left( {4.80\;{\rm{s}}} \right)\\d = 1679.57\;{\rm{m}}\end{aligned}\)

Evaluating the temperature during the winter day by using the its relation with velocity in winter

The speed of sound during winter day is calculated below:

\(\begin{aligned}{c}d = {v_2}{t_2}\\{v_2} = \frac{d}{{{t_2}}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{v_2} = \frac{{1679.57\;{\rm{m}}}}{{5.20\;{\rm{s}}}}\\{v_2} = 323\;{\rm{m/s}}\end{aligned}\)

The temperature on the winter day is calculated below:

\(\begin{aligned}{c}{v_2} = 331 + 0.61\left( {{T_2}} \right)\\{T_2} = \frac{{{v_2} - 331}}{{0.61}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{T_2} = \frac{{323\;{\rm{m/s}} - 331}}{{0.61}}\\{T_2} = - 13.11^\circ {\rm{C}}\end{aligned}\)

Hence, the temperature on the winter day is \( - 13.11^\circ {\rm{C}}\).