Sound intensity level to energy absorbed and power from loudspeaker

[dnl65078]

Homework Statement

a)A sound wave with an intensity level of 80.1 dB is incident on an eardrum of area 0.600 10-4 m2. Calculate the energy is absorbed by the eardrum in 4.00 minutes in microJ

b)The sound level 25.0 m from a loudspeaker is 69.0 dB. Calculate the rate at which sound energy is produced by the loudspeaker, assuming it to be an isotropic source.

Homework Equations

The Attempt at a Solution

We know tha the refernece intensity (I0) = 10-12W/m2
We know that sound intensity level is
β = 10dB log (I/Io)
Then 80.1dB = 10dB log (I/Io)
8.01 = log (I/Io)
Now finding I then substitue in the equation
I =P/A
= E/AΔt
Then the area covered by the sound is
A = 0.600 10-4 m2.
Then energy is absorbed by the eardrum in 4.00 minutes ( i.e 4*60s = 240s) in μJ.
E = IAΔt

I got 1.44x10^-14 (which ended up being wrong))b)
We know tha the refernece intensity (I0) = 10-12W/m2
We know that sound intensity level is
β = 10dB log (I/Io)
Then 69dB = 10dB log (I/Io)
6.9 = log (I/Io)
Now finding I then substitue in the equation
I =P/A
Then the area covered by the sound is
A = 4π(25.0m)2
= 7850m2
Now the total power of the source is
P = IA in W Or J/s

I got 7.85x10^-9 please help. what am I doing wrong?

 

[CompuChip]

Your calculation in a) looks correct, although I got 1.47... x 10^-6 J. Did you round before your final answer? Also, how did you get a 10^-14, are you sure you did not go from J to MJ instead of the other way around?

I'll let you check b) again then, probably it's a similar mistake.