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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 \(3000.0 \mathrm{~K}\) if heat conduction along the supports is ignored?

Short Answer

Expert verified
The power input required to maintain the sphere at 3000 K can be found by calculating the values in the steps above

Step by step solution

01

Understand the variables

The emissivity \(e\) is given as 0.350, the radius \(r\) of the sphere is 1.50 cm, the temperature of the surroundings \(T_{surroundings}\) is 290 K and the temperature of the sphere \(T_{sphere}\) is 3000 K.
02

Calculate the Surface Area of the sphere

As the tungsten sphere is spherical the Surface Area \(A\) can be calculated using the formula \(A = 4 \pi r^2\). Here the radius is provided in cm which has to be converted to meter as \(1.50 cm = 0.015 m\), the surface area would then be \(A = 4 \pi (0.015 m)^2\).
03

Solve for Power

Substitute the values for \(e\), \(A\), \(T_{sphere}\), and \(T_{surroundings}\) into the Stefan-Boltzmann law to get the power. The Stefan-Boltzmann constant \(\sigma\) is \(5.67 \times 10^{-8}\) watts per square meter per kelvin to the 4th power. After substituting, compute the expression.

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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 an object's ability to emit thermal radiation compared to a perfect emitter, known as a blackbody, at the same temperature. Ranging from 0 to 1, emissivity quantifies how effectively an object’s surface radiates energy. An emissivity value of 1 signifies that the object is an ideal emitter, radiating heat energy as efficiently as it theoretically can.

For the tungsten sphere in the exercise, the given emissivity is 0.350. This indicates that tungsten does not emit thermal radiation as efficiently as a blackbody. The lower emissivity value means less radiation is emitted for a given temperature compared to an ideal blackbody. This characteristic is vital when calculating the amount of power needed to maintain the sphere at a specific temperature, since emitted power must compensate for this inefficiency.
Thermal Radiation
Thermal radiation is the process by which the surface of an object emits energy in the form of electromagnetic waves due to the object's temperature. The fundamental principle governing thermal radiation is the Stefan-Boltzmann law, which states that the power radiated by an object is directly proportional to the fourth power of its absolute temperature.

In the context of the problem, a sphere made of tungsten aims to be kept at 3000 K, and we are interested in the amount of thermal radiation it emits. The law is mathematically expressed as: \[ P = e \times A \times \text{\(\sigma\)} \times (T_{\text{\( sphere \)}}^{4} - T_{\text{\( surroundings \)}}^{4}) \] where \( P \) denotes the radiated power, \( e \) the emissivity, \( A \) the surface area, \( \sigma \) the Stefan-Boltzmann constant, and \( T \) the temperature.
Surface Area Calculation
Calculating the surface area of a three-dimensional object is crucial when determining how much thermal radiation the object will emit. For our exercise, we are considering a sphere, which has a surface area given by the formula: \[ A = 4 \pi r^2 \] where \( r \) is the radius of the sphere. For accurate calculations, it's essential to use consistent units, so converting the radius from centimeters to meters is necessary because the Stefan-Boltzmann constant is defined using meters.

Once the radius is converted to meters, we apply the formula to find the surface area, which in turn is used to calculate the power necessary to maintain the temperature according to the Stefan-Boltzmann law. Such computations are fundamental in various applications, including engineering and environmental science, where understanding how objects interact with thermal radiation is essential.

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Most popular questions from this chapter

A machinist bores a hole of diameter \(1.35 \mathrm{~cm}\) in a steel plate that is at \(25.0^{\circ} \mathrm{C}\). What is the cross-sectional area of the hole (a) at \(25.0^{\circ} \mathrm{C}\) and \((\mathrm{b})\) when the temperature of the plate is increased to \(175^{\circ} \mathrm{C} ?\) Assume that the coefficient of linear expansion remains constant over this temperature range.

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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?

An asteroid with a diameter of \(10 \mathrm{~km}\) and a mass of \(2.60 \times 10^{15} \mathrm{~kg}\) impacts the earth at a speed of \(32.0 \mathrm{~km} / \mathrm{s},\) landing in the Pacific Ocean. If \(1.00 \%\) of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of \(\left.10.0^{\circ} \mathrm{C}\right),\) what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10^{15} \mathrm{~kg} .\)

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