Question

Compare rates of loss of heat by the body at temperature 527°C and 127°C. Temperature of surrounding is 27°C.

Answer 1

Given: 

T₁ = 527°C = 527 + 273 = 800 K

T₂ = 127°C = 127 + 273 = 400 K

T₀ = 27°C = 27 + 273 = 300 K

To find: `((dQ)/dt)_1/(((dQ)/dt)_2)` = ?

Formula:

`((dQ)/dt)_1=sigmaAe[T_1^4-T_0^4]`

`((dQ)/dt)_2=sigmaAe[T_2^4-T_0^4]`

∴ The ratio of the rates of loss of heat,

`((dQ)/dt)_1/(((dQ)/dt)_2) = (T_1^4-T_0^4)/(T_2^4-T_0^4) = ((800)^4-(300)^4)/((400)^4-(300)^4)`

`=(4096-81)/(256-81)=4015/175` 

`((dQ)/dt)_1/(((dQ)/dt)_2) = 22.94`