/*! 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 4 The temperature of an object res... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The temperature of an object resembling a blackbody is raised from \(200 \mathrm{K}\) to \(2000 \mathrm{K} .\) By how much does the amount of energy it radiates increase?

Short Answer

Expert verified
The energy radiated increases by 9999 times.

Step by step solution

01

Understanding the Blackbody Radiation

A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation. It radiates energy according to the Stefan-Boltzmann law, which states that the energy radiated per unit area is proportional to the fourth power of the blackbody's temperature.
02

Applying the Stefan-Boltzmann Law

The Stefan-Boltzmann Law is given by the formula \( E = \sigma T^4 \), where \( E \) is the energy radiated per unit area, \sigma is the Stefan-Boltzmann constant \( 5.67 \times 10^{-8} \ \mathrm{W} \mathrm{m}^{-2} \mathrm{K}^{-4} \), and \( T \) is the absolute temperature in Kelvin.
03

Calculating Energy at Initial Temperature

Initially, the temperature \( T_1\) is 200 K. The energy radiated per unit area at this temperature is given by \( E_1 = \sigma T_1^4 = (5.67 \times 10^{-8}) \times (200)^4 \).
04

Calculating Energy at Final Temperature

The final temperature \( T_2 \) is 2000 K. The energy radiated per unit area at this temperature is \( E_2 = \sigma T_2^4 = (5.67 \times 10^{-8}) \times (2000)^4 \).
05

Finding the Increase in Radiated Energy

The increase in energy radiated is \( E_2 - E_1 \). Using the formula for both energies, calculate the difference. Simplify \((2000/200)^4 = 10^4 = 10000\) to show \( E_2 = 10000 \times E_1\). The increase is therefore \( 10000 \times E_1 - E_1 = 9999 \times E_1\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in physics that describes how energy is radiated from a blackbody. This law is crucial for understanding how temperature affects the amount of energy an object emits. According to the law, the energy radiated per unit area of a blackbody (\(E\)) is directly proportional to the fourth power of the object's absolute temperature (\(T\)). The formula for the Stefan-Boltzmann Law is:
  • \(E = \sigma T^4\)
Here, \(\sigma\) represents the Stefan-Boltzmann constant, which is approximately \(5.67 \times 10^{-8} \, \mathrm{W} \, \mathrm{m}^{-2} \, \mathrm{K}^{-4}\). By applying this law, we can calculate how much energy a blackbody emits from its surface at any given temperature. It illustrates that a small increase in temperature leads to a significant increase in the energy radiated, as energy scales with the fourth power of temperature. This principle is key when evaluating how objects like stars emit energy.
Temperature and Energy Relationship
Temperature and energy are profoundly connected through blackbody radiation principles. As the temperature of an object increases, the amount of energy it emits also increases. This relationship is clearly quantified by the Stefan-Boltzmann Law. When considering a change in temperature from, say, 200 K to 2000 K, the energy increase can be substantial. Using the formula \(E = \sigma T^4\), when the temperature increases tenfold, the energy increases by a factor of \(10000\). This highlights the sensitivity of emitted energy to temperature changes. This dramatic rise means that even modest increases in temperature for objects resembling a blackbody can result in substantial changes in their energy output. Understanding this relationship helps us interpret various physical phenomena, such as why high-temperature stars emit more energy than cooler stars.
Electromagnetic Radiation
Electromagnetic radiation is the transfer of energy through waves or particles at varying frequencies and energies. A blackbody, an idealized object used in thermal physics, absorbs all incoming electromagnetic radiation and emits energy purely based on its temperature. This emitted radiation spans a spectrum that can include visible light, infrared, ultraviolet, and beyond. As an object's temperature rises, it emits more energy at higher frequencies. This is why hotter objects often appear brighter or can radiate energy in forms such as X-rays or gamma rays. In the context of the given exercise, increasing the temperature of a blackbody significantly changes the amount and type of electromagnetic radiation it emits. Understanding electromagnetic radiation allows us to explore diverse concepts, from the warmth of sunlight to technologies like microwave ovens, which rely on specific frequencies to heat food. Each part of the electromagnetic spectrum carries unique properties, but they all stem from the same fundamental phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A gas laser has a Fabry-Perot cavity of length \(40 \mathrm{cm}\). The index of refraction of the gas is \(1.0 .\) Operating at \(600 \mathrm{nm}\), determine the mode number, that is, the number of half-cycles fitting within the cavity.

The 488.0 -nm line from an argon ion laser is Doppler broadened to \(2.7 \times 10^{9} \mathrm{Hz}\). Given that the laser's mirrors are \(1.0 \mathrm{m}\) apart, determine the approximate number of longitudinal modes. Assume the index of refraction of the gas is 1.0.

A \(50.0-\mathrm{cm}^{3}\) chamber is filled with argon gas to a pressure of \(20.3 \mathrm{Pa}\) at a temperature of \(0^{\circ} \mathrm{C}\). All but a negligible number of these atoms are initially in their ground states. A flash tube surrounding the sample energizes \(1.0 \%\) of the atoms into the same excited state having a mean life of \(1.4 \times 10^{-8}\) s. What is the maximum rate at which photons are subsequently emitted by the gas (of course, it falls off with time \() ?\) Assume both that spontaneous emission is the only mechanism at work and that the medium is an ideal gas.

Suppose we have a 100 -W yellow lightbulb \((550 \mathrm{nm}) 100 \mathrm{m}\) away from a 3 -cm-diameter shuttered aperture. Assuming the bulb to have a \(2.5 \%\) conversion to radiant power, how many photons will pass through the aperture if the shutter is opened for \(\frac{1}{1000}\) s?

Given that a ruby laser operating at \(694.3 \mathrm{nm}\) has a frequency bandwidth of \(50 \mathrm{MHz}\), what is the corresponding linewidth?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.