Question

compare the rates of loss of heat by black body at 827°C and at 427°C. if the temperature of surrounding is 27°C​

Answer 1

Answer:

             T₁ = 827⁰C +273 = 1100K

             T₂ = 427⁰C +273 = 700K

             Tₓ = 27°C  +273  = 300K

according to stefan boltzmann law of radiation.

        $\frac{dq}{dt}$       = σΑε(T₁⁴-Tₓ⁴)

     ($\frac{dq}{dt} /\frac{dq}{dt}$ )  = (T₁⁴-Tₓ⁴)/T₂⁴-Tₓ⁴ = [1100⁴-300⁴]/[700⁴-300⁴]

                  = 6.276

So now we can compare the rates as 6.276:1 .  and this is a rates of loss of heat by black body.

Answer 2

Explanation:

here is the answer of and pls before doing check the image question properly