The temperature of a lamp filament is 2100 K | vedclass

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(B) The power radiated by a body is given by the Stefan-Boltzmann Law: $P = e \sigma A T^4$.Here,$e = 0.453$,$\sigma = 5.67 	imes 10^{-8} \ W/m^2K^4$

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) The power radiated by a body is given by the Stefan-Boltzmann Law: $P = e \sigma A T^4$.Here,$e = 0.453$,$\sigma = 5.67 	imes 10^{-8} \ W/m^2K^4$,$A = 4 	imes 10^{-4} \ m^2$,and $T = 2100 \ K$.Substituting the values:$P = 0.453 	imes (5.67 	imes 10^{-8}) 	imes (4 	imes 10^{-4}) 	imes (2100)^4$.$P = 0.453 	imes 5.67 	imes 4 	imes 10^{-12} 	imes (2.1 	imes 10^3)^4$.$P = 10.26792 	imes 10^{-12} 	imes 19.4481 	imes 10^{12}$.$P \approx 199.73 \ W$.Rounding to the nearest integer,the power is $200 \ W$.

The temperature of a lamp filament is $2100 \ K$ and its surface area is $4 \times 10^{-4} \ m^2$. If the emissivity of the filament is $0.453$,then the power of the lamp will be ....... $W$.

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