# Change in radiation with temperature

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## Homework Statement

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

P=σεAT^4

## The Attempt at a Solution

dP/P=4dT/T dT=5 when T is 100 initially. Lets 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
Homework Helper
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You could just make the ratio of the initial and final values, since the differential change is not asked.

Last edited:
Gold Member
You could just make the ratio of the initial and final values, since the differential change is not asked.

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
Homework Helper
Gold Member
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.

Gold Member
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
Homework Helper
Gold Member
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

Gold Member
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