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A black body radiates heat energy at the rate of 2 x 10 Js" m' at a temperature of 127° C. The temperature

of the black body at which the rate of heat radiation is 32 x 105 Js'1 m2

is​

Respuesta :

MathPhys

Answer:

527 ° C

Explanation:

For black body radiation, the power emitted per unit area is directly proportional to the absolute temperature raised to the fourth power:

j = k T⁴

First, convert 127° C to Kelvins:

127° C + 273.15 = 400.15 K

Now find the constant of variation:

2×10⁵ = k (400.15)⁴

k = 7.80×10⁻⁶

Finally, solve for the temperature at the new rate of radiation:

32×10⁵ = (7.80×10⁻⁶) T⁴

T = 800.3

The temperature is 800 K, or about 527 ° C.

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