/*! 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 74 The emissivity of tungsten is \(... [FREE SOLUTION] | 91Ó°ÊÓ

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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 515 W.

Step by step solution

01

Understand the Stefan-Boltzmann Law

The Stefan-Boltzmann Law is used to calculate the power radiated by an object based on its temperature. It is given by the formula: \( P = \varepsilon \cdot \sigma \cdot A \cdot (T^4 - T_0^4) \), where \( P \) is the power, \( \varepsilon \) is the emissivity, \( A \) is the surface area of the sphere, \( T \) is the temperature of the sphere, \( T_0 \) is the temperature of the surroundings, and \( \sigma \) is the Stefan-Boltzmann constant, \( 5.67 \times 10^{-8} \, \mathrm{W/m^2 \cdot K^4} \).
02

Calculate the Surface Area of the Sphere

The surface area \( A \) of a sphere is given by the formula \( A = 4 \pi r^2 \), where \( r \) is the radius of the sphere. Given that the radius \( r = 1.50 \, \mathrm{cm} = 0.015 \, \mathrm{m} \), we calculate the surface area as follows:\[ A = 4 \pi (0.015)^2 \approx 2.827 \times 10^{-3} \, \mathrm{m^2} \]
03

Substitute the Values into the Stefan-Boltzmann Equation

Now substitute the given values into the Stefan-Boltzmann Law:- \( \varepsilon = 0.350 \)- \( \sigma = 5.67 \times 10^{-8} \, \mathrm{W/m^2 \cdot K^4} \)- \( A = 2.827 \times 10^{-3} \, \mathrm{m^2} \)- \( T = 3000 \, \mathrm{K} \)- \( T_0 = 290.0 \, \mathrm{K} \)The formula becomes:\[ P = 0.350 \cdot 5.67 \times 10^{-8} \cdot 2.827 \times 10^{-3} \cdot ((3000)^4 - (290)^4) \]
04

Calculate the Power Output

Performing the calculation:First, calculate the temperatures raised to the fourth power:- \( 3000^4 = 8.1 \times 10^{13} \, \mathrm{K^4} \)- \( 290^4 = 7.11 \times 10^{9} \, \mathrm{K^4} \)Now, substitute these into the formula:\[ P = 0.350 \cdot 5.67 \times 10^{-8} \cdot 2.827 \times 10^{-3} \cdot ((8.1 \times 10^{13}) - (7.11 \times 10^{9})) \]This gives:\[ P \approx 0.350 \cdot 5.67 \times 10^{-8} \cdot 2.827 \times 10^{-3} \cdot 8.0992929 \times 10^{13} \]\[ P \approx 515.37 \, \mathrm{W} \]
05

Conclusion and Round the Answer

The calculated power input required to maintain the sphere at 3000.0 K is approximately \( 515.37 \, \mathrm{W} \). For practical purposes, we round this to \( 515 \, \mathrm{W} \).

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Key Concepts

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

Emissivity
Emissivity is a measure of how effectively an object emits thermal radiation. It is represented by the symbol \( \varepsilon \) and ranges from 0 to 1.
A perfect blackbody, which is an ideal emitter, has an emissivity of 1, while an object that does not emit thermal radiation at all has an emissivity of 0.
In practical terms, most materials have emissivity values between these two extremes.
The Stefan-Boltzmann Law takes emissivity into account when calculating the power radiated by an object. The formula, \( P = \varepsilon \cdot \sigma \cdot A \cdot (T^4 - T_0^4) \), includes emissivity as a factor that modifies the ideal blackbody radiation to match the actual material properties.
In this exercise, the emissivity of tungsten is given as \(0.350\), which means that it emits 35% as much thermal radiation as an ideal blackbody at the same temperature. Understanding emissivity helps in predicting how much energy an object will radiate under given circumstances.
Surface Area of a Sphere
To find out how much thermal radiation a sphere emits, you need to know its surface area.
The formula for the surface area \( A \) of a sphere is \( A = 4 \pi r^2 \), where \( r \) is the radius of the sphere.
This formula is derived from geometric principles and is essential for understanding how energy emission scales with the size of the object.
In the given problem, the sphere's radius is 1.50 cm, which we convert to meters (\( r = 0.015 \, \mathrm{m} \)) for consistency with SI units.
Plugging this into the surface area formula, we get:\[ A = 4 \pi (0.015)^2 \approx 2.827 \times 10^{-3} \, \mathrm{m^2} \]
This surface area is used in the Stefan-Boltzmann Law to calculate the power emitted by the sphere, showing the essential role geometry plays in the physics of radiation.
Power Input Calculation
To keep a tungsten sphere at a stable high temperature, one must calculate the power input needed to balance the thermal radiation it emits. The calculation involves several quantitative aspects.
The Stefan-Boltzmann Law is our tool, expressed as \( P = \varepsilon \cdot \sigma \cdot A \cdot (T^4 - T_0^4) \).
Let's break down the process:
  • Use the sphere's emissivity (\( \varepsilon = 0.350 \)) and surface area (\( A = 2.827 \times 10^{-3} \, \mathrm{m^2} \))
  • Apply the Stefan-Boltzmann constant \( \sigma = 5.67 \times 10^{-8} \, \mathrm{W/m^2 \cdot K^4} \)
  • Insert temperatures \( T = 3000 \, \mathrm{K} \) (sphere) and \( T_0 = 290 \, \mathrm{K} \) (surroundings)
These values lead to the equation for power \( P \), which can be solved to find:\[ P \approx 515.37 \, \mathrm{W} \]
Power input equal to this value is required to maintain the given conditions, balancing the thermal radiation emitted by the sphere.
Thermal Radiation
Thermal radiation is a crucial concept in understanding how objects emit energy in the form of electromagnetic waves. Any object at a temperature above absolute zero emits thermal radiation due to the movement of its charged particles.
The key law governing this process is the Stefan-Boltzmann Law, which not only depends on temperature but also factors in the material's emissivity and surface area.
When calculating the power radiated by the tungsten sphere, the difference in temperatures accounted in the formula \( (T^4 - T_0^4) \) represents the directional flow of energy - from the hot object to the cooler surroundings.
Understanding thermal radiation helps in various applications, from designing heating systems to explaining natural phenomena like how Earth emits energy into space.
Temperature Difference
The role of temperature difference in thermal radiation can't be overstated. It dictates the net flow of energy between objects, influencing the power output.
In the Stefan-Boltzmann equation, the term \( (T^4 - T_0^4) \) represents the difference between the temperatures of the emitting body and its surroundings, raised to the fourth power.
This non-linear relationship means that even small changes in temperature can drastically alter the radiation power.
In our scenario, the sphere's temperature is 3000 K, while the surroundings are at 290 K. Despite this seeming small relativity in difference when simply subtracting, the fourth power raises the impact of these temperatures significantly.
Thus, understanding how the steep sensitivity of thermal radiation to temperature changes is vital for accurate predictions and adjustments in practical applications, like maintaining consistent heat levels in the tungsten sphere.

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Most popular questions from this chapter

A vessel whose walls are thermally insulated contains 2.40 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice, all at a temperature of \(0.0^{\circ} \mathrm{C}\) . The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to \(28.0^{\circ} \mathrm{C}\) ? You can ignore the heat transferred to the container.

A nail driven into a board increases in temperature. If we assume that 60\(\%\) of the kinetic energy delivered by a 1.80-kg hammer with a speed of 7.80 \(\mathrm{m} / \mathrm{s}\) is transformed into heat that flows into the nail and does not flow out, what is the temperature increase of an \(8.00-\mathrm{g}\) aluminum nail after it is struck ten times?

A U.S. penny has a diameter of 1.9000 \(\mathrm{cm}\) at \(20.0^{\circ} \mathrm{C} .\) The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10^{-5} \mathrm{K}^{-1} .\) What would its diameter be on a hot day in Death Valley \(\left(48.0^{\circ} \mathrm{C}\right) ?\) On a cold night in the mountains of Greenland \(\left(-53^{\circ} \mathrm{C}\right) ?\)

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50 \(\mathrm{kW} / \mathrm{m}^{2} .\) The distance from the earth to the sun is \(1.50 \times 10^{11} \mathrm{m},\) and the radius of the sun is \(6.96 \times 10^{8} \mathrm{m}\) . (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

Bicycling on a Warm Day. If the air temperature is the same as the temperature of your skin (about \(30^{\circ} \mathrm{C} ),\) your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 70 -kg person's body produces energy at a rate of about 500 \(\mathrm{W}\) due to metabolism, 80\(\%\) of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\) . (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750 -mL bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is 1.0 \(\mathrm{kg.}\)

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