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 ∝ $\sqrt{T}$              (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),

$\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$

$T_{2}$ = $T_{2} = (\frac{v_{2}}{v_{1}})^{2}T_{1}$

$T_{2} = (\frac{340}{349})^{2}{293}$ = 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