A thin, totally absorbing sheet of mass m, face area A, and specificheat${\mathit{c}}_{\mathbf{s}}$ isfully illuminated by a perpendicular beam of a plane electromagnetic wave. The magnitude of the maximum electric field of the wave is${\mathit{E}}_{\mathbf{m}}$.What is the rate$\frac{\mathbf{d}\mathbf{T}}{\mathbf{d}\mathbf{t}}$at which the sheet’s temperature increases due to the absorption of the wave?

• Mass of the sheet is$m$
• Face area of the sheet is$A$
• Specific heat of the sheet is$c_s$
• Maximum electric field of wave is$E_m$

By using the concept of specific heat, the intensity of the electromagnetic wave, and the rate of flow of energy, we will find the rate of increase in temperature.

Formulae:

The absorbed heat by the body due to specific heat,$Q=c_{s}m\Delta T\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

The intensity of the wave due to electric field,$I=\frac{1}{2c{\mu }_0}{E}_m^2\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

The rate of energy absorption by the body,$P=\frac{dE}{dt}=IA\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)$

Substituting the intensity value of equation (2) in equation (3), the rate of energy can be given as follows:

$\frac{dE}{dt}=\frac{1}{2C{\mu }_0}{E}_m^{2}A$

Inthegiven problem, electric field energy is transferred as heat energy. Thus, we can write that

$\begin{array}{c}\frac{dQ}{dt}=\frac{dE}{dt}\\ =\frac{1}{2C{\mu }_0}{E}_m^{2}A\cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)\end{array}$

$\begin{array}{c}\frac{dQ}{dt}=c_{s}m\frac{dT}{dt}\\ \frac{dT}{dt}=\frac{1}{c_{s}m}\frac{dQ}{dt}\\ =\frac{1}{c_{s}m}\frac{1}{2c{\mu }_0}{E}_m^{2}A\left(\text{fromequation(4)}\right)\\ =\frac{1}{2c{\mu }_0{c}_{s}m}{E}_m^{2}A\end{array}$

Hence, the rate at which the sheet’s temperature increases due to the absorption of the wave is.$\frac{1}{2c{\mu }_0{c}_{s}m}{E}_m^{2}A$