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Find the emitted power per square meter of peak intensity for a 3000 __.

a. Kelvin source
b. Watt bulb
c. Lumen flashlight
d. Nanometer wavelength

User Inspire
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1 Answer

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Final answer:

The emitted power per square meter of peak intensity for a 3000 Kelvin source is 1.365 × 105 W/m2.

Step-by-step explanation:

To find the emitted power per square meter of peak intensity for a 3000 Kelvin source, you need to use the Stefan-Boltzmann Law. This law states that the power emitted by a perfect black body is directly proportional to the fourth power of its temperature.

The equation for the power emitted per square meter by a black body is given by P = σT4, where P is the power, σ is the Stefan-Boltzmann constant (5.67 × 10-8 W/m2K4), and T is the temperature in Kelvin.

Substituting the given temperature of 3000 Kelvin into the equation, we can calculate the emitted power per square meter.

Power emitted per square meter = 1.365 × 105 W/m2

User Kah
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