Question

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?

Short answer

The rate at which the sheet’s temperature increases due to the absorption of the waveis.$\frac{E_m^{2}A}{2c{\mu }_{0}m{c}_s}$

Step-by-step solution

Given data 

• 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$E_m$

Understanding the concept of the specific heat

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,$P=\frac{dE}{dt}=IA\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)$

Calculation of the rate of temperature of the sheet 

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

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

Now, the rate of heat absorbed by the body can be given using equation (1) as follows:

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

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