# Change in radiation with temperature

• Krushnaraj Pandya
In summary, the conversation discusses how to calculate the increment in heat energy radiated when the temperature of a hot body is raised by 5%. Different methods are suggested, including using the ratio of initial and final values and using the error method. It is determined that the error method may provide an approximation of the error, but the ratio method gives exact values. The accuracy of the approximation depends on the size of the error.
Krushnaraj Pandya
Gold Member

## 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. 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?

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##?

Last edited:
stockzahn said:
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

Krushnaraj Pandya said:
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.

stockzahn said:
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

Krushnaraj Pandya said:
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

stockzahn said:
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

## 1. How does temperature affect radiation?

Temperature has a direct impact on the amount of radiation emitted by an object. As the temperature increases, the amount of radiation also increases. This is because higher temperatures cause atoms and molecules to vibrate at a faster rate, which in turn leads to the emission of more radiation.

## 2. What is the relationship between temperature and radiation intensity?

The relationship between temperature and radiation intensity is directly proportional. This means that as the temperature increases, the radiation intensity also increases. Similarly, as the temperature decreases, the radiation intensity decreases as well.

## 3. Can temperature affect the wavelength of radiation?

Yes, temperature can affect the wavelength of radiation. As the temperature increases, the wavelength of radiation decreases, and vice versa. This is because higher temperatures cause atoms and molecules to vibrate at a faster rate, resulting in shorter wavelengths being emitted.

## 4. How does the type of material affect the relationship between temperature and radiation?

The type of material does not affect the relationship between temperature and radiation. The relationship remains the same regardless of the material, as it is based on the properties of atoms and molecules, which are consistent across all materials.

## 5. How can changes in radiation with temperature be measured?

Changes in radiation with temperature can be measured using a device called a pyrometer. A pyrometer measures the amount of infrared radiation emitted by an object and can then calculate the temperature based on this information. Other methods of measurement include using a spectrometer or thermal imaging camera.

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