Question

The speed of sound in air is proportional to the square root of the absolute temperature. If the speed of sound is 349 m/s when the air temperature is 20 °C, what is the temperature of the air when the speed of sound is 340 m/s? Give your answer in °C, K, °F, and R (Rankine).

Answer 1

Given:

Let the speed of sound be represented by 'v' then

v ∝
`√(T)` (1)

`v_(1)` = 349 m/s

`v_(2)` = 340 m/s

`T_(1)` = 20°C = 273+20 = 293 K

Formulae used:

1) °C = K + 273

2) K = °C - 273

3) °F = 1.8°C + 32

4) °R = °F + 459.67

Solution:

From eqn (1),

`(v_(1))/(v_(2)) = \sqrt{(T_(1))/(T_(2))}`

`T_(2)` =
`T_(2) = ((v_(2))/(v_(1)))^(2)T_(1)`

`T_(2) = ((340)/(349))^(2){293}` = 278.08 K

Now, Usinf formula (1), (2), (3) and (4) respectively, we get

1) T = 293 K

2) T = 293 -278.8 = 5.08°C

3) T = 1.8(5.08) + 32=41.14°F

4) T = 41.14 + 459.67 = 500.81°R