Question

The temperature of a blackbody is increased by 50% then percentage increase in rate of radiation is approximately

Answer 1

Answer:81/16times

Explanation:E is directly proportional to T^4

T1=t

T2=t+1r/2=3t/2

Applying the relation

T2=81/16 E1

Answer 2

Hey Dear,

◆ Answer -

Percentage increase in rate of radiation is approximately 406 %.

● Explanation -

According to Stephens law, rate of emission of radiation by a black body is directly proportional to fourth power of temperature.

R ∝ T^4

When temperature is increased by 50% i.e. 150%,

R'/R = (T'/T)^4

R'/R = (150T/100T)^4

R'/R = (1.5)^4

R'/R = 5.0625

R' = 506.25% R

Therefore, percentages increase in rate of radiation is approximately 506 - 100 = 406 %.

Hope this helps you.