The temperature is observed to decrease from 32$^{\circ}$C to 23$^{\circ}$C during the evening of a very hot day.What is the relative change in the speed of sound over this time interval?
A) -15%
B) -28%
C) -1.5%
D) -3%

Question

Quizplus · Accepted Answer

The speed of sound is directly proportional to the square root of the absolute temperature.
Let's assume the initial absolute temperature (T1) as follows:
T1 = (32 + 273) K = 305 K
And, the final absolute temperature (T2) as follows:
T2 = (23 + 273) K = 296 K
The relative change in speed of sound (Δv/v) can be calculated by using the following formula:
Δv/v = (1/2) * (ΔT/T)
Where ΔT = T2 - T1 and T = (T1 + T2)/2
Substituting the given values in the above equation, we get:
Δv/v = (1/2) * [(296 - 305)/(300.5)] = -0.015 or -1.5%
Therefore, the answer is C) -1.5%.

The Temperature Is Observed to Decrease from 32 ∘^{\circ}∘ C to 23

The Temperature Is Observed to Decrease from 32 $^{\circ}$ C to 23