A window whose area is `2m^2` opens on street where the street noise result in an intensity level at the window of 60 dB. How much acoustic power enters the window via sound waves. Now if an acoustic absorber is fitted at the window, how much energy from street will it collect in 5 h?

Explanation

Loudness level is given as
`L = (10log)(I/I_(0))`
`(10log) (I)/(I_(0)) = 60`
`(I)/(I_(0)) = 10^(6)`
`I = (10^(-12)) xx 10^(6)) = 10^(-6)W//m^(2) = 1muW//m^(2)`
And we have power of sound is p = IS
`1xx10^(-6) xx 2 = 2muW`
Thus energy is given as
` E = p xx t
= 2xx 10^(-6) xx 5 xx 60 xx 60 = 36 xx 10^(-3) J`