Change in radiation with temperature

[Krushnaraj Pandya]

Homework Statement

What would be the increment in heat energy radiated when the temperature of a hot body is raised by 5%?

Homework Equations

P=σεAT^4

The Attempt at a Solution

dP/P=4dT/T dT=5 when T is 100 initially. Let's assume P was also 100 initially for convenience, therefore dP should be 20, but the answer given is 21.55%, where am I wrong and what's the correct way to approach this?

 

[stockzahn]

You could just make the ratio of the initial and final values, since the differential change is not asked.

Edit: Shouldn't it read $\frac{P+dP}{P}=\left(\frac{T+dT}{T}\right)^4$?

 

[Krushnaraj Pandya]

You could just make the ratio of the initial and final values, since the differential change is not asked.

Edit: Shouldn't it read $\frac{P+dP}{P}=\left(\frac{T+dT}{T}\right)^4$?

I got what you wrote, but using errors method can't we write dP/P = 4dT/T as well. For example in measuring the change in time period of pendulum on changing length. Since T∝L^0.5 we write ΔT/T = ΔL/2L

 

[stockzahn]

I got what you wrote, but using errors method can't we write dP/P = 4dT/T as well. For example in measuring the change in time period of pendulum on changing length. Since T∝L^0.5 we write ΔT/T = ΔL/2L

Is suppose an error af 5 % cannot be considered as small anymore. Try it with smaller errors (like $10^{-5}$), then your method works very well.

 

[Krushnaraj Pandya]

Is suppose an error af 5 % cannot be considered as small anymore. Try it with smaller errors (like $10^{-5}$), then your method works very well.

So this method should give me an approximation of the error in any case, correct? (Time is invaluable in multiple choice exams)
While taking the ratio gives the exact values

 

[stockzahn]

So this method should give me an approximation of the error in any case, correct? (Time is invaluable in multiple choice exams)
While taking the ratio gives the exact values

But the approximation gets worse with increasing error, in your case (fourth power):

error - exact - approximation
$10^{-6}$ - 0.000004 - 0.000004
$10^{-5}$ - 0.000040001 - 0.00004
$10^{-4}$ - 0.00040006 - 0.0004
$10^{-3}$ - 0.004006004 - 0.004
$10^{-2}$ - 0.04060401 - 0.04
$5\cdot10^{-2}$ - 0.21550625 - 0.2

It depends on your application if the approximation is sufficient

 

[Krushnaraj Pandya]

But the approximation gets worse with increasing error, in your case (fourth power):

error - exact - approximation
$10^{-6}$ - 0.000004 - 0.000004
$10^{-5}$ - 0.000040001 - 0.00004
$10^{-4}$ - 0.00040006 - 0.0004
$10^{-3}$ - 0.004006004 - 0.004
$10^{-2}$ - 0.04060401 - 0.04
$5\cdot10^{-2}$ - 0.21550625 - 0.2

It depends on your application if the approximation is sufficient

Oh, alright. Thank you very much