Question

Compare the rates of emission of heat by a blackbody maintained at 627°C and at 127°C, if the black bodies are surrounded by an enclosure at 27°C. What would be the ratio of their rates of loss of heat? 

Answer 1

Given:

T₀ = 27°C = 27 + 273 = 300 K
T₁ = 627°C = 627 + 273 = 900 K
T₂ = 127°C = 127 + 273 = 400 K

To find: Ratio of rate of loss of heat (R1 : R2)

Formula: R = `"dQ"/"dt" = "e"σ"A"("T"^4 - "T"_0^4)`

Calculation: 

From formula,

R₁ = `("dQ"/"dt")_1 = "e"σ"A"("T"_1^4 - "T"_0^4)` .....(1)

R₂ = `("dQ"/"dt")_2 = "e"σ"A"("T"_2^4 - "T"_0^4)` .....(2)

Dividing equation (1) by equation (2),

`"R"_1/"R"_2 = ("T"_1^4 - "T"_0^4)/("T"_2^2 - "T"_0^4) = (900^4 - 300^4)/(400^4 - 300^4)`

= `(9^4 - 3^4)/(4^4 - 3^4)`

= `6480/175`

= antilog {log (6480) − log (175)}

= antilog {3.8116 − 2.2430}

= antilog {1.5686}

= 37.03

The ratio of the rate of energy radiated is 37.03:1.