# Chapter 12 - Problems - Page 463: 14

The energy absorbed by the eardrum in 3.00 minutes is $1.08\times 10^{-6}~J$

#### Work Step by Step

We can find the intensity of the sound: $\beta = 10~log\frac{I}{I_0}$ $80.0 = 10~log\frac{I}{I_0}$ $8.0 = log\frac{I}{I_0}$ $10^{8.0} = \frac{I}{I_0}$ $I = (10^{8.0})~I_0$ $I = (10^{8.0})~(1.0\times 10^{-12}~W/m^2)$ $I = 1.0\times 10^{-4}~W/m^2$ We can find the power on the eardrum: $P = I~A$ $P = (1.0\times 10^{-4}~W/m^2)(0.600\times 10^{-4}~m^2)$ $P = 6.0\times 10^{-9}~W$ We can find the energy absorbed by the eardrum in 3.00 minutes: $E = P~t = (6.0\times 10^{-9}~W)(180~s) = 1.08\times 10^{-6}~J$ The energy absorbed by the eardrum in 3.00 minutes is $1.08\times 10^{-6}~J$

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