The emissivity of tungsten is 0.3500.350. A tungsten sphere with radius 1.501.50 cm is suspended within a large evacuated enclosure whose walls are at 290.0290.0 K. What power input is required to maintain the sphere at 3000.03000.0 K if heat conduction along the supports is ignored?

Ch 17: Temperature and Heat

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 17: Temperature and Heat

Problem 68

Chapter 17, Problem 68

The emissivity of tungsten is $0.350$ . A tungsten sphere with radius $1.50$ cm is suspended within a large evacuated enclosure whose walls are at $290.0$ K. What power input is required to maintain the sphere at $3000.0$ K if heat conduction along the supports is ignored?

Verified step by step guidance

Start by understanding the concept of thermal radiation. The power radiated by an object can be calculated using the Stefan-Boltzmann law, which states that the power radiated per unit area is proportional to the fourth power of the temperature.

The formula for the power radiated by a sphere is given by: $ P = \varepsilon \sigma A (T^4 - T_0^4) $, where $ \varepsilon $ is the emissivity, $ \sigma $ is the Stefan-Boltzmann constant $ 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 $, $ A $ is the surface area of the sphere, $ T $ is the temperature of the sphere, and $ T_0 $ is the temperature of the surroundings.

Calculate the surface area $ A $ of the sphere using the formula for the surface area of a sphere: $ A = 4 \pi r^2 $, where $ r $ is the radius of the sphere. Convert the radius from centimeters to meters before calculating.

Substitute the values into the Stefan-Boltzmann equation: $ P = 0.350 \times 5.67 \times 10^{-8} \times A \times ((3000)^4 - (290)^4) $.

Perform the calculations step by step, ensuring units are consistent, to find the power input required to maintain the sphere at 3000 K.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Emissivity

Emissivity is a measure of an object's ability to emit thermal radiation compared to a perfect black body. It ranges from 0 to 1, where 1 represents a perfect emitter. In this problem, tungsten has an emissivity of 0.350, indicating it emits 35% of the radiation a black body would at the same temperature.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power radiated by a black body is proportional to the fourth power of its temperature, given by P = εσAT^4, where ε is emissivity, σ is the Stefan-Boltzmann constant, A is the surface area, and T is the temperature. This law is crucial for calculating the power needed to maintain the sphere's temperature.

Surface Area of a Sphere

The surface area of a sphere is calculated using the formula A = 4πr^2, where r is the radius. For the tungsten sphere with a radius of 1.50 cm, this formula helps determine the area over which thermal radiation is emitted, essential for applying the Stefan-Boltzmann Law to find the required power input.

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