Light Absorption and Heat Transfer

[Mehmet Saygın]

The same intensity of blue (450-495 nm), green (495-570 nm), yellow (500-730 nm), red (620-730 nm) and white colored lights fall on identical yellow surfaces in the same environment. Which of these surfaces will be the warmest after a while?

[swansont]

What do you think the answer is, and why?

[Mehmet Saygın]

2 hours ago, swansont said:

What do you think the answer is, and why?

I am undecided between two options which are white light and blue light. If we think the visible spectrum,white light has a wider range which can be described between 380-730 nm. Basically, yellow surface reflects between 500-730 nm and absorbs between 380-500 nm. Blue light between 450-495 nm is also absorbed by yellow surface. The absorbed range is wider at white light, so the answer seems to be as white light at this perspective, but on the other hand, the light intensity is equal. Red,green and yellow lights are reflected from yellow surface,white light is absorbed partially, only all of blue light absorbed from yellow surface. The answer looks like blue light according to this idea. I am not sure which idea is the correct one.

Edited April 2, 2018 by Mehmet Saygın
Misspelling

[Sensei]

3 hours ago, Mehmet Saygın said:

yellow (500-730 nm)

yellow is between 570–590 nm..

ps. Excuse me for being too pedantic..

 

[Mehmet Saygın]

26 minutes ago, Sensei said:

yellow is between 570–590 nm..

ps. Excuse me for being too pedantic..

 

It does not matter ☺️

I know what do you mean but this 570-590 nm for pure yellow light. When I said 500-730 nm, our yellow light reflects all wavelength between green-red part of visible spectrum as like an ideal yellow object. I hope I can explain the key. ☺️

[swansont]

1 hour ago, Mehmet Saygın said:

I am undecided between two options which are white light and blue light. If we think the visible spectrum,white light has a wider range which can be described between 380-730 nm. Basically, yellow surface reflects between 500-730 nm and absorbs between 380-500 nm. Blue light between 450-495 nm is also absorbed by yellow surface. The absorbed range is wider at white light, so the answer seems to be as white light at this perspective, but on the other hand, the light intensity is equal. Red,green and yellow lights are reflected from yellow surface,white light is absorbed partially, only all of blue light absorbed from yellow surface. The answer looks like blue light according to this idea. I am not sure which idea is the correct one.

If all of the blue light is absorbed, and only some of the other colors are absorbed (including white), what does that tell you about how much energy is absorbed in a given amount of time? 

[Mehmet Saygın]

17 minutes ago, swansont said:

If all of the blue light is absorbed, and only some of the other colors are absorbed (including white), what does that tell you about how much energy is absorbed in a given amount of time? 

I think,I can not explain myself correctly ☺️

1) Blue light (450-495 nm) absorbed fully.

2) Green light (495-570 nm) reflected fully.

3) Yellow light (500-730 nm) reflected fully.

4) Red light (620-730 nm) reflected fully.

5) White light (380-730 nm) reflected (500-730 nm) and absorbed (380-500 nm).

In this case,the correct answer looks like as white light because it has biggest absorption part.

But the light intensity is equal for all lights, so only small part of white light is absorbed (380-500 nm), but blue light is absorbed fully.

I mean light intensity factor can be change the answer white light to blue light ?

 

 

[swansont]

31 minutes ago, Mehmet Saygın said:

I think,I can not explain myself correctly ☺️

1) Blue light (450-495 nm) absorbed fully.

2) Green light (495-570 nm) reflected fully.

3) Yellow light (500-730 nm) reflected fully.

4) Red light (620-730 nm) reflected fully.

5) White light (380-730 nm) reflected (500-730 nm) and absorbed (380-500 nm).

In this case,the correct answer looks like as white light because it has biggest absorption part.

But the light intensity is equal for all lights, so only small part of white light is absorbed (380-500 nm), but blue light is absorbed fully.

I mean light intensity factor can be change the answer white light to blue light ?

The problem tells us that the light sources all have the same intensity. So if we have (as an example) 100 W of light (per unit area), all 100 W of blue is absorbed. Something less than 100 W of white (or any color source) is absorbed, since it will reflect some of the light.

[MaximThibodeau]

 

 

In radiometry, irradiance is the radiant flux (power) received by a surface per unit area.
Irradiance is often called intensity because it has the same physical dimensions.

 

The Planck–Einstein relation connects the particular photon energy E with its associated wave frequency f:
E = h f

 

h = 6.626070040(81)×10−34 j*s

 

blue: 495 - 450 = 45
green = 570 - 495 = 75
yellow = 730 - 500 = 230
red = 730 - 620 = 110
white = 730 - 450 = 280

 

Power mean Energy/time could lead to:(depending on the scope of your school class?)

for blue: E = (hf2 - hf1)/45 because of the linearity
for white: you take the two triangles: but your data are not correct: yellow 730

 

I dont know, what is your school level?

Edited April 2, 2018 by MaximThibodeau

[swansont]

13 minutes ago, MaximThibodeau said:

 

 

In radiometry, irradiance is the radiant flux (power) received by a surface per unit area.
Irradiance is often called intensity because it has the same physical dimensions.

 

The Planck–Einstein relation connects the particular photon energy E with its associated wave frequency f:
E = h f

 

h = 6.626070040(81)×10−34 j*s

 

blue: 495 - 450 = 45
green = 570 - 495 = 75
yellow = 730 - 500 = 230
red = 730 - 620 = 110
white = 730 - 450 = 280

 

Power mean Energy/time could lead to:(depending on the scope of your school class?)

for blue: E = (hf2 - hf1)/45 because of the linearity
for white: you take the two triangles: but your data are not correct: yellow 730

 

I dont know, what is your school level?

How does any of this relate to the question being asked? Part of it is wrong, and none of it appears to help in answering the question — there's nothing that requires the use of Planck's constant, and there's no reason to be subtracting the wavelength. 

[MaximThibodeau]

3 minutes ago, swansont said:

there's nothing that requires the use of Planck's constant

Because, it is a more fondamental proof and help understanding the problem.

 

4 minutes ago, swansont said:

there's no reason to be subtracting the wavelength

To average it, by the Integral that could be reduce to a triangle by linearity

[Mehmet Saygın]

1 hour ago, MaximThibodeau said:

 

 

In radiometry, irradiance is the radiant flux (power) received by a surface per unit area.
Irradiance is often called intensity because it has the same physical dimensions.

 

The Planck–Einstein relation connects the particular photon energy E with its associated wave frequency f:
E = h f

 

h = 6.626070040(81)×10−34 j*s

 

blue: 495 - 450 = 45
green = 570 - 495 = 75
yellow = 730 - 500 = 230
red = 730 - 620 = 110
white = 730 - 450 = 280

 

Power mean Energy/time could lead to:(depending on the scope of your school class?)

for blue: E = (hf2 - hf1)/45 because of the linearity
for white: you take the two triangles: but your data are not correct: yellow 730

 

I dont know, what is your school level?

Thanks for a detailed information, I am a highschool student.

The range of yellow is described as 500-730 nm,because the yellow light reflects between all green-red part of visible spectrum in this question. It means that the light is not monochromatic pure yellow light (570-590 nm) as you think.

[Sensei]

25 minutes ago, Mehmet Saygın said:

The range of yellow is described as 500-730 nm,because the yellow light reflects between all green-red part of visible spectrum in this question. It means that the light is not monochromatic pure yellow light (570-590 nm) as you think.

If "yellow" is mixture of green and red (like in CRT/LED screens), which just pretends only to human eyes it's yellow, you should say so since the beginning..

 

[swansont]

1 hour ago, MaximThibodeau said:

Because, it is a more fondamental proof and help understanding the problem.

 

To average it, by the Integral that could be reduce to a triangle by linearity

 

15 minutes ago, Sensei said:

If "yellow" is mixture of green and red (like in CRT/LED screens), which just pretends only to human eyes it's yellow, you should say so since the beginning..

C'mon people. Let's keep in mind that this is HW help, and anyone asking questions can only go with what they have been taught. Answers need to be given that conform to the material that's presented. If it's a simple "heat absorbed raises temperature" problem (as it is here) then it's inappropriate to bring quantum mechanics into the discussion, or to be answering questions other than what was asked, or to kibbutz on the information that's given in the problem, over which the student has no control.