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Writer's picturekeshprad

PhysicsBowl 2018 Q50


50) For this question, we want to find how the rate of energy radiation changes as the radius of the sphere is doubled. The rate of energy radiation is essentially the power of the object.

Where P is power, I is intensity, and A is area

This is the general form of the power of an object. Now using the Stefan-Boltzmann law of blackbody radiation, we can rewrite this equation for this situation.

Where P is power, I is intensity, A is area, e is emissivity, σ is the Stefan-Boltzmann constant, T is temperature, and A is area

Our object is a sphere, so as our radius changes, the surface area will change as well. Specifically, we know that the surface area is proportional to r^2.

Where P is power, e is emissivity, σ is the Stefan-Boltzmann constant, T is temperature, and r is radius

When we double radius r, we are multiplying the equation by a factor of 4.


Answer: C

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