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clarified for which medium the reference intensity is relevant
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honeste_vivere
honeste_vivere
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First, I went into great detail on the propagation of sound in different media in the following answer: https://physics.stackexchange.com/a/266046/59023.

Sound

Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 20 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2 for water.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2 under water.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.

Answer

I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

First, I went into great detail on the propagation of sound in different media in the following answer: https://physics.stackexchange.com/a/266046/59023.

Sound

Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 20 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.

Answer

I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

First, I went into great detail on the propagation of sound in different media in the following answer: https://physics.stackexchange.com/a/266046/59023.

Sound

Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 20 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2 for water.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2 under water.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.

Answer

I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

Fixed the reference pressure of air
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honeste_vivere
honeste_vivere
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First, I went into great detail on the propagation of sound in different media in the following answer: https://physics.stackexchange.com/a/266046/59023.

Sound

Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 1020 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.

Answer

I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

First, I went into great detail on the propagation of sound in different media in the following answer: https://physics.stackexchange.com/a/266046/59023.

Sound

Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 10 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.

Answer

I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

First, I went into great detail on the propagation of sound in different media in the following answer: https://physics.stackexchange.com/a/266046/59023.

Sound

Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 20 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.

Answer

I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/
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First, I went into great detail on the propagation of sound in different media in the following answer: http://physics.stackexchange.com/a/266046/59023https://physics.stackexchange.com/a/266046/59023.

Sound

Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 10 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.

Answer

I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

First, I went into great detail on the propagation of sound in different media in the following answer: http://physics.stackexchange.com/a/266046/59023.

Sound

Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 10 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.

Answer

I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

First, I went into great detail on the propagation of sound in different media in the following answer: https://physics.stackexchange.com/a/266046/59023.

Sound

Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 10 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2.

We can then use: $$ I = I_{o} \ 10^{L/10} $$ where $I$ is intensity (in W/m2) and $L$ is intensity (in dB). If we use the intensity I show above and $L$ ~ 230 dB, then $I \sim 6.74 \times 10^{4}$ W/m2.

If we use your 1 m2 area and $\Delta t$ ~ 1 ms, then the total energy would be ~67.4 J.

Answer

I am not sure if I should be using this 61 dB offset in my calculation above. Is the energy of the "click" still 100 MJ?

No, it is much lower at ~67.4 J. The $\sim 10^{11}$ W you found is rather extreme when you consider that it corresponds to peak electrical power consumption of France. Though whales are rather enormous mammals and a ~10,000 calorie diet does correspond to $\sim 10^{7}$ J of energy (i.e., diet of an Olympic athlete, which is probably much less than a sperm whale), that would still be a lot of power to produce.

typos
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endolith
endolith
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honeste_vivere
honeste_vivere
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