/*! 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 76 The emissivity of tungsten is \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The emissivity of tungsten is \(0.350 .\) A tungsten sphere with radius 1.50 \(\mathrm{cm}\) is suspended within a large evacuated enclosure whose walls are at 290.0 \(\mathrm{K}\) . What power input is required to maintain the sphere at a temperature of 3000.0 \(\mathrm{K}\) if heat conduction along the supports is neglected?

Short Answer

Expert verified
The required power input is approximately 1840 W.

Step by step solution

01

Calculate the Surface Area of the Sphere

To find the surface area of the sphere, use the formula for the surface area of a sphere: \( A = 4\pi r^2 \). The radius \( r \) is given as \(1.50 \text{ cm} = 0.015 \text{ m} \). Therefore, \( A = 4\pi (0.015)^2 \approx 2.827\times 10^{-3} \text{ m}^2 \).
02

Use the Stefan-Boltzmann Law

The power radiated by a black body is given by the Stefan-Boltzmann Law: \( P = \epsilon \sigma A (T^4 - T_0^4) \). Here, \( \epsilon = 0.350 \) is the emissivity of tungsten, \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \) is the Stefan-Boltzmann constant, \( A \approx 2.827\times 10^{-3} \text{ m}^2 \) is the surface area, \( T = 3000 \text{ K} \) is the temperature of the sphere, and \( T_0 = 290 \text{ K} \) is the temperature of the enclosure walls.
03

Substitute Values into the Stefan-Boltzmann Equation

Substitute the known values into the Stefan-Boltzmann equation: \[ P = 0.350 \times 5.67 \times 10^{-8} \times 2.827\times 10^{-3} \times (3000^4 - 290^4) \].
04

Calculate the Power Radiated

First, calculate \( (3000^4 - 290^4) \). This results in \( 3000^4 \approx 8.1 \times 10^{13} \) and \( 290^4 \approx 7.1 \times 10^{9} \), so \( 3000^4 - 290^4 \approx 8.0999 \times 10^{13} \). Now the power is \( P = 0.350 \times 5.67 \times 10^{-8} \times 2.827\times 10^{-3} \times 8.0999 \times 10^{13} \).
05

Final Calculation

Perform the final multiplication to find the power: \( P \approx 0.350 \times 5.67 \times 10^{-8} \times 2.827\times 10^{-3} \times 8.0999 \times 10^{13} \approx 1.84 \times 10^{3} \text{ W} \).

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 cornerstone principle in thermal radiation. It describes how much power a black body object emits based on its temperature. This law states that the power radiated by an object is proportional to the fourth power of its absolute temperature.
Feel like Einstein for a moment with the formula: \[P = \epsilon \sigma A (T^4 - T_0^4)\]Here’s what each variable means:
  • \(P\): The power radiated in watts (W).
  • \(\epsilon\): Emissivity of the material, indicating how efficient the object is at emitting thermal radiation.
  • \(\sigma\): The Stefan-Boltzmann constant, a fixed number at approximately \(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\).
  • \(A\): Surface area of the object in square meters (\(m^2\)).
  • \(T\): Absolute temperature of the object in Kelvin (K).
  • \(T_0\): Absolute temperature of the surrounding.
This law helps determine how much energy a surface radiates purely based on its temperature.
This is crucial for understanding how much energy needs to be supplied to maintain the desired temperature when external conditions vary.
Emissivity
Emissivity is a measure of a material's ability to emit thermal radiation. Think of it as the efficiency rating for radiating energy.
If the emissivity is 1, the object is a perfect emitter, known as a "black body."
Real-world objects have emissivities less than 1, showing they do not emit as efficiently as an ideal black body. For example:
  • A black body has an emissivity of \(1\), meaning it radiates energy perfectly.
  • Tungsten, like in our example, has an emissivity of \(0.35\). This means it radiates 35% of the energy it would as a perfect black body.
Emissivity depends on several factors:
  • Material type: Metals, non-metals, and ceramics all have different emissivities.
  • Surface roughness: Polished surfaces emit less than rough surfaces.
  • Wavelength and temperature: These factors can change emissivity values.
Remember, higher emissivity means better heat radiation.
This directly affects how much energy needs to be input, especially when maintaining temperatures far from the ambient conditions.
Black Body Radiation
Black body radiation pertains to an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. This means a perfect black body is a perfect emitter, crucial in thermal radiation as it represents the maximum energy radiated for a given temperature.
In essence, black bodies can:
  • Absorb all radiation, making them appear black when at a temperature below visible light emission.
  • Emit radiation solely based on their temperature, according to the Stefan-Boltzmann Law.
You can think of them as the perfect template or benchmark for studying thermal radiation.
In practical terms, even though no perfect black bodies exist in nature, many objects approximate this behavior closely enough for theoretical models.
Ideal black bodies fully support understanding emissivity and radiation models since they offer a baseline comparison for real-world objects that do not emit perfectly.
Heat Transfer Calculation
Heat transfer calculations are central to determining the power input needed to maintain a specified temperature.
In the exercise, it’s all about finding out how much power or energy is necessary.
Here's how to approach such problems:
  • Start by determining the surface area of the object since this influences the radiation amount.
  • Apply the Stefan-Boltzmann Law to calculate the radiated power, while considering the emissivity and temperatures given.
  • Subtract the environment’s effect (\(T_0\)) to see the net power loss.
By calculating accurately:
  • You ensure you can precisely adjust the power needed to keep the system in thermal balance.
  • This shows how the interactions between object temperature, environment, and material properties govern overall thermal behavior.
Remember, thorough understanding aids in optimizing systems that require specific thermal conditions, from small devices to large industrial applications.

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) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

A worker pours 1.250 kg of molten lead at a temperature of \(365.0^{\circ} \mathrm{C}\) into 0.5000 \(\mathrm{kg}\) of water at a temperature of \(75.00^{\circ} \mathrm{C}\) in an insulated bucket of negligible mass. Assuming no heat loss to the surroundings, calculate the mass of lead and water remaining in the bucket when the materials have reached thermal equilibrium.

Find the Celsius temperatures corresponding to (a) a winter night im Seattle \(\left(41.0^{\circ} \mathrm{F}\right) ;(\mathrm{b})\) a hot summer day in Palm Springs \(\left(107.0^{\circ} \mathrm{F}\right) ;(\mathrm{c})\) a cold winter day in northern Manitoba \(\left(-18.0^{\circ} \mathrm{F}\right)\) .

Hot Air in a Physics Lecture. (a) Atypical student listening attentively to a physics lecture has a heat output of 100 \(\mathrm{W}\) . How much heat energy does a class of 90 physics students release into a lecture hall over the course of a 50 -min lecture? (b) Assume that all the heat energy in part (a) is transferred to the 3200 \(\mathrm{m}^{3}\) of air in the room. The air has specific heat capacity 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and density 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) . If none of the heat escapes and the air conditioning system is off, how much will the temperature of the air in the room rise during the 50 -min lecture? (c) If the class is taking an exam, the heat output per student rises to 280 \(\mathrm{W}\) . What is the temperature rise during 50 \(\mathrm{min}\) in this case?

Heat Loss During Breathing. In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C}\) , what amount of heat is needed to warm to body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the 0.50 \(\mathrm{L}\) of air exchanged with each breath? Assume that the specific heat of air is 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that 1.0 \(\mathrm{L}\) of air has mass \(1.3 \times 10^{-3} \mathrm{kg}\) . (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

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.