Refer again to question 6 in which they are using sonar with a loudness of 235 d

Question

Refer again to question 6 in which they are using sonar with a loudness of 235 dB as measured at 1 m. Say an unfortunate fish is swimming 0.5 m in front of the sound source. The fish’s body tissue has a density of 1.06 g/cm3, and the speed of sound in its tissue is 1.6 X 105 cm/s. The sea water has a density of 1.0 g/cm3, and the speed of sound in this water is still 1.4 X 105 cm/s. Treat the fish as a sphere with a mass of 5 kg and a specific heat of 0.83 cal/g°C. The fish’s body tissues absorb all of the sound energy that enters the body. At what rate would its temperature rise due to the absorbed sound energy if it sheds none of that heat to the environment? Ignore the void created by the air bladder.

Explanation / Answer

Impedance of a tissue defined =density of tissue * velocity of sound in that tissue.

Z2=1.06 g/cm3* 1.6 X 105 cm/s=169600 gm/cm.s

Acoustic Impedance (Z) is similar to electrial resistance.

Intensity given 235 db or  amount of energy flowing through a unit cross-sectional area of a beam each second

Z1=1.00*1.4*10^5= 140000 gm/cm.s

Now reflection coefficeint R=((Z2-Z1)/(Z2+Z1))=((169600-140000)/(169600+140000))*100=9.56%

R= the amount of energy reflected as a percentage of the original energy

Intensity at 0.5 m will be 940 db i.e on the surface of fish

After reflection, the energy transmitited will be =940-940*0.0956= 850.1360 dbm

will be equal to 1.03*10^85 watt or joule per second or Q/t

Q=mSdT or we can Q/t=mSdT/t or dT/t=Q/t/mS=(1.03*10^85)/(5*3475.042496)= 5.9280e+080 *C/s

specific heat unit converted

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