Question

A blackbody at 327 °C, when suspended in a black enclosure at 27 °C, loses heat at a certain rate. Find the temperature of the body at which its rate of loss of heat by radiation will be half of the above rate. Assume that the other conditions remain unchanged. 

Answer 1

T₁ = 273 + 327 = 600 K, T₀ = 273 + 27 = 300 K,

(dQ/dt)₂ = `((dQ//dt)_1)/2`

∴ `((dQ//dt)_1)/(dQ//dt)_2 = 2`

Let T₂ be the required temperature.

`(dQ)/dt = σ A (T^4 - T_0^4)`

∴ `((dQ//dt)_1)/((dQ//dt)_2)`

= `(T_1^4 - T_0^4)/((T_2^4 - T_0^4))`

when the other conditions remain unchanged.

∴ 2 = `(T_1^4 - T_0^4)/(T_2^4 - T_0^4)`

∴ `2 T_2^4 - 2 T_0^4`

= `T_1^4 - T_0^4`

∴ `T_2^4 = (T_1^4 + T_0^4)/2`

= `((600)^4 + (300)^4)/2`

= 688.5 × 10⁸

∴ T₂ = 512.2 K