Question

A black metal foil is warmed by radiation from a small sphere at temp t at a distance d it is found that the power received by the foil is p

Answer 1

As ,  

Þ P ∝ T ^ 4 /  d ^ 2

Þ P1 / P2 = [ T1 / T2 ] ^ 4  x   [ d2 / d1 ] ^ 2

Þ P / P2 = [ T / 2T ] ^ 2   x  [ 2d / d ] ^ 2

  Power received per second     = 1 / 4  Þ P2 = 4 P

Answer 2

Answer:

Solution:

Energy Received by the foil per second is, power $P =(T^4 -T_{0}^{4} )$

Thus, $P\quad \alpha\quad T^{4}$

Also the energy received per second is inversely proportional to the square of distance,

     ${ P\quad \alpha \left( \frac { 1 }{ d } \right) }^{ 2 }$

 Now,    ${ P\quad \alpha \quad T }^{ \left( \frac { 4 }{ d } \right) \overset { 2 }{ \underset} }$

$P\left( \frac { 1 }{ P }\right) ^{ 2 }=\left( \frac { { T }_{ 1 } }{ { T }_{ 2 } } \right) ^{ 4 } \times \quad \left( \frac { { d }_{ 2 } }{ { d }_{ 1 } } \right) ^{ 2 }$

$\frac { P }{ { P }_{ 2 } } \quad =\quad \left( \frac { T }{ 2T } \right) ^{ 4 }\times \quad \left( \frac { 2d }{ d } \right) ^{ 2 }$ (since both the temperature and distance are doubled.)

             $P = \frac{1}{4} \times P_2$

             $P_2 = 4P$