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The tungsten filament of a light bulb has an operating temperature of about 2 100 K. If the emitting area of the filament is 1.0 cm2, and its emissivity is 0.68, what is the power output of the light bulb? (σ = 5.67 × 10−8W/m2⋅K4)
a. 100 Wb. 75 Wc. 60 Wd. 40 W

Respuesta :

Answer:

75 W

Explanation:

[tex]T[/tex] = temperature of the filament = 2100 K

[tex]A[/tex] = Emitting area of the filament = 1 cm² = 10⁻⁴ m²

[tex]e[/tex] = Emissivity = 0.68

[tex]\sigma[/tex] = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the light bulb is given as

[tex]P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (0.68) (10^{-4}) (2100)^{4}\\P = 75 W[/tex]

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