Answer
(a) Energy is absorbed by the water at a rate of $914~W$
(b) $I = 1.04\times 10^5~W/m^2$
(c) $E_{rms} = 6260~V/m$
$B_{rms} = 2.09\times 10^{-5}~T$
Work Step by Step
(a) We can find the energy required to raise the temperature of the water:
$E = m~C~\Delta~T$
$E = (0.350~kg)(4180~J/kg~K)(75.0~K)$
$E = 109,725~J$
We can find the rate at which energy is absorbed:
$P = \frac{E}{t} = \frac{109,725~J}{120~s} = 914~W$
Energy is absorbed by the water at a rate of $914~W$
(b) We can find the average intensity:
$I = \frac{P}{A} = \frac{914~J}{88.0\times 10^{-4}~m^2} = 1.04\times 10^5~W/m^2$
(c) We can find $E_{rms}$:
$E_{rms}^2 = \frac{I}{c~\epsilon_0}$
$E_{rms} = \sqrt{\frac{I}{c~\epsilon_0}}$
$E_{rms} = \sqrt{\frac{1.04\times 10^5~W/m^2}{(3.0\times 10^8~m/s)~(8.85\times 10^{-12}~C^2/N~m^2)}}$
$E_{rms} = 6260~V/m$
We can find $B_{rms}$:
$B_{rms} = \frac{E_{rms}}{c}$
$B_{rms} = \frac{6260~V/m}{3.0\times 10^8~m/s}$
$B_{rms} = 2.09\times 10^{-5}~T$
