# The speed of an ultrasonic sound frequency 45 kHz in air is 342 m/s. What is the air temperture?

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You should use the equation that relates the speed of sound and temepratures such that:

`v_(sound) = 331m/s + 0.6T` .

The problem provides the value of speed of sound, hence, substituting 342 m/s for `v_(sound)` yields:

`342 m/s = 331 m/s + 0.6T =gt 0.6T = 342 m/s- 331 m/s`

`0.6T = 11 m/s =gt T = 11/0.6 =gt T = 11/(6/10) `

`T = 11*10/6 =gt T = 11*5/3 = 18.3^oC`

**Hence, evaluating the temperature under the given conditions yields `T = 18.3^oC.` **

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