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The filament temperature of a light bulb is 2 000 K when the bulb delivers 40 W of power.

If its emissivity remains constant, what power is delivered when the filament temperature is 2 500 K?

a. 105 W

b. 62 W

c. 98 W

d. 50 W

Respuesta :

Answer:

c. 98 W

Explanation:

[tex]P_{i}[/tex] = initial power delivered by bulb at initial temperature = 40 W

[tex]T_{i}[/tex] = initial temperature of filament = 2000 K

Now the temperature of filament of the bulb rises

[tex]P_{f}[/tex] =Final power delivered by bulb at higher temperature

[tex]T_{f}[/tex] = Final higher temperature of filament = 2500 K

As per Stefan's law, the power delivered by the bulb depends directly on fourth power of the temperature. hence we can write

[tex]\frac{P_{f}}{P_{i}} = \frac{T_{f}^{4} }{T_{i}^{4}} \\\frac{P_{f}}{40} = \frac{2500^{4} }{2000^{4}}\\\\P_{f} = (2.44) (40)\\P_{f} = 98 W[/tex]

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