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

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610당신이 사용하는 게 아니라요?교과서 변경

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Young & Freedman Calc 14th Edition

Ch 17: Temperature and Heat

문제 68

17장, 문제 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?

검증된 단계별 안내

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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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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