Two lightbulbs have cylindrical filaments much greater in length than in diameter. The evacuated bulbs are identical except that one operate

Question

Two lightbulbs have cylindrical filaments much greater in length than in diameter. The evacuated bulbs are identical except that one operates at a filament temperature of 2 1008C and the other operates at 2 0008C. (a) Find the ratio of the power emitted by the hotter lightbulb to that emitted by the cooler lightbulb. (b) With the bulbs operating at the same respective temperatures, the cooler lightbulb is to be altered by making its filament thicker so that it emits the same power as the hotter one. By what factor should the radius of this filament be increased

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Thu Cúc 4 years 2021-07-22T00:02:52+00:00 1 Answers 27 views 0

Answers ( )

  1. Thunguyet
    0
    2021-07-22T00:04:33+00:00
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    Answer:

    a

    \frac{P_1}{P_2} = 1.188 \ m

    b

    Z = 1.08995

    Explanation:

    From the question we are told that

    The filament temperature of the first bulb is T_1 = 2100 ^o C = 2100 + 273 = 2373 \ K

    The filament temperature of the second bulb is T_3 = 2 0008^o C = 2000 + 273 = 2273 \ K

    Generally according to Stefan-Boltzmann law the power emitted by first bulb is mathematically represented as

    \frac{P}{A} = \sigma T^4_1

    Here \sigma is the Stefan-Boltzmann with value \sigma = 5.67*10^{-8} \ W . m ^{-2}. K ^{-4}

    So

    \frac{P_1}{A_1} = (5.67*10^{-8}) (2373)^4

    => P_1 = 1797935 A_1

    Generally according to Stefan-Boltzmann law the power emitted by second bulb is mathematically represented as

    \frac{P_2}{A_2} = \sigma T^4_2

    \frac{P_2}{A_2} = (5.67*10^{-8}) * (2273)^4

    => P_1 = 1513494 A_2

    Given that the two bulbs are identical we have that

    A_1 = A_2 = A

    So

    The ration is mathematically represented as

    \frac{P_1}{P_2} = \frac{1797935* A}{1513494 *A}

    => \frac{P_1}{P_2} = 1.188 \ m

    Generally the area is mathematically represented as

    A = \pi r^2

    Recall that

         A_1 = A_2 = A

    =>   \pi r^2_1 = \pi r^2_2 = \pi r^2

    =>   r^2_1 = r^2_2 = r^2

    Now if the radius of the cooler bulb is increase by a factor Z (i.e Z * r )then the area of the cooler bulb A_2 [\tex] becomes [tex] A_2 = (Zr)^2

    =>     A_2 = Z^2*r^2

    Here Z is the factor by which it is made thicker

    So  For  \frac{P_1}{P_2} = \frac{1797935* A}{1513494 *A}

    1.188 = \frac{1797935* r^2}{1513494 *Z^2*r^2}

    => Z = 1.08995

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