Answer to Question #163721 in Mechanical Engineering for ali

Question

Two rectangles SO by 50 cm are placed perpendic J arly with a common e Ti 1000 K while surface two balance with 2 large surround ling room at 300 heat lost from the surface at 1000 K to surface number 2.

Answer 1

Answer

We know Stephen law

E=$\sigma T^4$

Where

$\sigma =5.67\times10^{-8}J/T$

Now for given temperature

$E_1=\sigma T_1^4$

$=5.67\times10^{-8}\times(1000) ^4$

$=56700W/m^2$

$E_2=\sigma T_2^4$

$=5.67\times10^{-8}\times(300) ^4$

$=45027W/m^2$

Therefore

Temperature of insulated surface

is

$T_2=(\frac{E_2}{\sigma}) ^{0.25}$

Putting the value of E₂ and Stephen constant

So

$T_2=(\frac{45027}{5.67\times10^{-8}}) ^{0.25}=566.378K$