4
$\begingroup$

Firewood has energy content of about 15 MJ/kg for hardwoods.

You throw one of these hardwood logs onto a fire hot enough that it burns with, say, 50% efficiency. How much of that energy will be released as sound?

$\endgroup$
6
$\begingroup$

Let's make the following assumptions:

  • One needs to add $1$ log/$30\,\rm min$ to keep the fire burning;
  • One log weights approximately $1\,\rm kg$.

Under these conditions, with your $50\%$ efficiency estimation, the power consumed by the fire is

$$P = \frac{15 \cdot 10^6\,\rm J \cdot 0.5}{1800\,\rm s} \approx 4.2 \cdot 10^3\,\rm W$$

which is consistent with the estimated heat output of a wooden stove.

I don't have a fireplace handy, sadly, and it's hard to find calibrated recordings of fire online. One can however, find recordings of fire. It is a common sound texture example. So I played a fire recording found on this page and adjusted the volume to match what I thought sounded like a natural fire. I used an FFT analyzer and found it peaked at roughly $40\,\rm dB$ SPL$^\dagger$.

I subsequently scaled the signal to obtain the same peak in the FFT and computed its power. I obtained $5 \cdot 10^{-9}\,\rm W$. Now, this underestimates the sound power generated by the fire, since it is actually based on the sound picked up by the microphone only, where the sound is radiated in 3D. With the assumption of a quarter-spherical sound field, and an admittedly loose interpretation of the inverse-square law, one can estimate that the sound pressure at the source was twice as high, and thus the power is underestimated by a factor of $4$. Thus let's say the fire has an approximate sound power output of $2 \cdot 10^{-8}\,\rm W$.

Here is a graph of the 'calibrated' waveform: enter image description here

So, my estimated answer would be that the fraction of the power that goes into sound is $~4 \cdot 10^{-12}$. And I could be wrong by several orders of magnitude and this ratio would still be minuscule.

Notes

$^\dagger$ SPL stands for sound pressure level, and $\mathrm{dB\,SPL}$ are defined as $20 \cdot \log_{10}\left(\frac{P_\mathrm{rms}}{20\,\rm \mu Pa}\right)$.

$\endgroup$
  • 3
    $\begingroup$ +1, and nice detective work! We're lucky that the most of the energy is heat and light; it would be pretty bad news if throwing a log on a fire leveled everything in a 20-block radius. $\endgroup$ – John Feminella Nov 5 '14 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.