Answer
(a) $0.0032J$
(b) $0.0081J$
Work Step by Step
(a) We know that
$I=I_{\circ}(10^{\frac{B}{10dB}})$
$I=(10^{-12}W/m^2)(10^{\frac{950dB}{10dB}})$
This simplifies to:
$I=0.00316W/m^2$
Now the energy absorbed by the eardrum for $4$ hours is given as
$E_4=Power\times time$
We plug in the known values to obtain:
$E_4=IA\times t$
$\implies E=I(\frac{\pi d^2}{4})\times t$
We plug in the known values to obtain:
$E_4=(0.00316W/m^2)\pi (\frac{9.5\times 10^{-3}}{2})^2(4h)(\frac{3600s}{1h})$
$\implies E_4=0.0032J$
(b) We know that
$I=I_{\circ}(10^{\frac{\beta}{{10dB}}})$
We plug in the known values to obtain:
$I=(10^{-12}W/m^2)(10^{\frac{105dB}{10dB}})$
$I=0.1316W/m^2$
Now the energy absorbed by the eardrum for 1 hour is given as
$E_1=IA\times t$
$E_1=I(\pi \frac{d^2}{4})\times t$
We plug in the known values to obtain:
$E_1=(0.0316W/m^2)(\pi \frac{(9.5\times 10^{-3})^2}{4})(1h)(36005)$
$\implies E_1=0.0081J$
