Question #163721

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.

Expert's answer

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 E_{2} and Stephen constant

So

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

Learn more about our help with Assignments: Mechanical Engineering

## Comments

## Leave a comment