/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 96 A thin, totally absorbing sheet ... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin, totally absorbing sheet of mass \(m\), face area \(A\), and specific heat \(c_{s}\) is fully illuminated by a perpendicular beam of a plane electromagnetic wave. The magnitude of the maximum electric field of the wave is \(E_{m}\). What is the rate \(d T / d t\) at which the sheet's temperature increases due to the absorption of the wave?

Short Answer

Expert verified
\(\frac{dT}{dt} = \frac{c \varepsilon_0 A}{4m c_s} E_m^2\)

Step by step solution

01

Understand the relationship between intensity and electric field

The intensity \(I\) of an electromagnetic wave is related to the rms value of its electric field \(E\) by the equation:\[ I = \frac{c \varepsilon_0}{2} E^2 \]where \(c\) is the speed of light in vacuum and \(\varepsilon_0\) is the permittivity of free space. The maximum electric field \(E_m\) is converted to rms value by the relation \(E = \frac{E_m}{\sqrt{2}}\).
02

Substitute the electric field relation into the intensity equation

Substitute \(E = \frac{E_m}{\sqrt{2}}\) into the intensity equation:\[ I = \frac{c \varepsilon_0}{2} \left(\frac{E_m}{\sqrt{2}}\right)^2 = \frac{c \varepsilon_0}{4} E_m^2 \]
03

Determine the power absorbed by the sheet

Since the sheet is totally absorbing, the power \(P\) absorbed by the sheet is given by:\[ P = I \times A = \frac{c \varepsilon_0 A}{4} E_m^2 \]
04

Relate absorbed power to energy and temperature change

The energy \(Q\) absorbed per unit time (power) can be related to the temperature change using specific heat \(c_s\):\[ P = \frac{dQ}{dt} = mc_s \frac{dT}{dt} \]where \(m\) is the mass of the sheet.
05

Solve for the rate of temperature increase

Equating the expressions for power and utilizing energy absorption, we can solve for \(\frac{dT}{dt}\):\[ \frac{c \varepsilon_0 A}{4} E_m^2 = mc_s \frac{dT}{dt} \]\[ \frac{dT}{dt} = \frac{c \varepsilon_0 A}{4m c_s} E_m^2 \]
06

Final expression for the rate of temperature increase

The final expression for the rate at which the sheet's temperature increases is:\[ \frac{dT}{dt} = \frac{c \varepsilon_0 A}{4m c_s} E_m^2 \]This equation gives the rate of temperature increase based on the properties of the wave and the sheet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intensity of Electromagnetic Waves
When discussing electromagnetic waves, one key concept is intensity, which represents the power per unit area carried by the wave. It's closely linked to the electric field of the wave, specifically the root mean square (RMS) value. The intensity \(I\) of an electromagnetic wave relates to the electric field \(E\) through the equation:
  • \( I = \frac{c \varepsilon_0}{2} E^2 \)
Here, \(c\) denotes the speed of light, and \(\varepsilon_0\) indicates the permittivity of free space. These constants are essential in understanding the propagation of electromagnetic waves in a vacuum. The maximum electric field, \(E_m\), can be transformed into an RMS value \(E\) by the relation:
  • \( E = \frac{E_m}{\sqrt{2}} \)
This transformation allows us to calculate the intensity based on the peak electric field magnitude. By substituting \(E\) into the intensity formula, we find that the intensity is directly proportional to the square of the maximum electric field. This relationship is critical in determining how much energy the wave imparts to a surface, as we see with the absorbing sheet.
Specific Heat Capacity
Specific heat capacity, often symbolized as \(c_s\), is a physical property describing how much energy is needed to change the temperature of a substance. It's usually expressed as the amount of energy required to raise the temperature of 1 kilogram of a material by 1 degree Celsius.
  • Formula: \( Q = mc_s \Delta T \)
Here, \(Q\) represents the heat energy added, \(m\) is the mass of the material, \(c_s\) is the specific heat capacity, and \(\Delta T\) is the temperature change.
Specific heat capacity varies between different materials, meaning some substances require more energy to increase their temperature compared to others. For our absorbing sheet, the specific heat capacity \(c_s\) plays a crucial role in determining how the absorbed wave energy translates into temperature change.
Understanding specific heat capacity helps us see the connection between energy input and temperature change, which is crucial for predicting thermal responses in materials exposed to electromagnetic waves.
Temperature Change in Materials
When materials absorb energy from electromagnetic waves, like in our example with the absorbing sheet, their temperature increases. The rate \(\frac{dT}{dt}\) at which the temperature of the material changes depends on several factors:
  • The power of the absorbed energy
  • The specific heat capacity of the material
  • The mass of the material
The relationship can be described by rearranging the formula for energy \(Q\), expressed as \(Q = mc_s \Delta T\), into a rate form:
  • \( \frac{dQ}{dt} = mc_s \frac{dT}{dt} \)
This equation links the absorbed energy rate \(\frac{dQ}{dt}\) to the rate of temperature change. For the absorbing sheet fully illuminated by the electromagnetic wave, the temperature increases at a rate given by:
  • \( \frac{dT}{dt} = \frac{c \varepsilon_0 A}{4m c_s} E_m^2 \)
This formula illustrates how the intensity, specific heat capacity, and mass of the material interact to determine how quickly the material's temperature will increase. Understanding this balance is key to addressing problems involving heat exchange and energy absorption in various applications.

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Most popular questions from this chapter

At a beach the light is generally partially polarized due to reflections off sand and water. At a particular beach on a particular day near sundown, the horizontal component of the electric field vector is \(2.3\) times the vertical component. A standing sunbather puts on polarizing sunglasses; the glasses eliminate the horizontal field component. (a) What fraction of the light intensity received before the glasses were put on now reaches the sunbather's eyes? (b) The sunbather, still wearing the glasses, lies on his side. What fraction of the light intensity received before the glasses were put on now reaches his eyes?

A beam of intensity \(I\) reflects from a long, totally reflecting cylinder of radius \(R ;\) the beam is perpendicular to the central axis of the cylinder and has a diameter larger than \(2 R\). What is the beam's force per unit length on the cylinder?

Calculate the (a) upper and (b) lower limit of the Brewster angle for white light incident on fused quartz. Assume that the wavelength limits of the light are 400 and \(700 \mathrm{~nm}\).

The average intensity of the solar radiation that strikes normally on a surface just outside Earth's atmosphere is \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). (a) What radiation pressure \(p_{r}\) is exerted on this surface, assuming complete absorption? (b) For comparison, find the ratio of \(p_{r}\) to Earth's sea-level atmospheric pressure, which is \(1.0 \times 10^{5} \mathrm{~Pa}\).

A beam of polarized light is sent into a system of two polarizing sheets. Relative to the polarization direction of that incident light, the polarizing directions of the sheets are at angles \(\theta\) for the first sheet and \(90^{\circ}\) for the second sheet. If \(0.10\) of the incident intensity is transmitted by the two sheets, what is \(\theta\) ?

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