Effect of temperature on speed of sound [closed]

Formula for calculating speed of sound in dry air is

$$V(t)=V(0)+0.61t$$

The temperature here is always taken in Celsius .Why don't we use kelvin?

closed as unclear what you're asking by ACuriousMind♦, user36790, Gert, CuriousOne, Qmechanic♦Feb 27 '16 at 0:53

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• It is a useful approximation for those who work in degree Celsius. en.wikipedia.org/wiki/… – Farcher Feb 25 '16 at 15:56
• ...because Kelvin would have a different formula (obtained by $t\mapsto t + 273.15$)? I don't know what your question is. – ACuriousMind Feb 25 '16 at 16:00
• It's an approximate scaling equation for small changes in temperature near zero Celsius. The temperature actually scales like the square root of the absolute temperature. – Bill N Feb 26 '16 at 21:57

$$v = \sqrt{\gamma RT}$$

Suppose we want to calculate the speed near some reference temperature $T_0$, i.e. the temperature is $T = T_0 + \delta T$ where $\delta T$ is small. We rewrite the equation for the velocity as:

$$v = \sqrt{\gamma R(T_0 + \delta T}$$

and then rearrange this to:

$$v = \sqrt{\gamma R T_0} \left(1 + \frac{\delta T}{T_0}\right)^{1/2}$$

Then because $\delta T \ll T_0$ we can expand the square root using a binomial expansion to get:

$$v \approx \sqrt{\gamma R T_0} \left(1 + \frac{1}{2}\frac{\delta T}{T_0}\right)$$

But the term $\sqrt{\gamma R T_0}$ is just the velocity at the temperature $T_0$ so our equation becomes:

$$v \approx v(T_0) + \frac{1}{2}\sqrt{\frac{\gamma R}{T_0}}\delta T$$

and that's how we get the approximate equation you cite. The particular case when $T$ is given in Celcius just comes from taking $T_0 = 273.15$K so there is nothing special about it - it's just a convenient choice of $T_0$.

• I think you're missing the molar mass in the denominator. – Bill N Feb 26 '16 at 21:59