Asked by GenChem
I am wondering if my work makes sense for the following problem:
Warm objects emit electromagnetic radiation in the infra-red region. Heat lamps employ this principle to generate infra-red radiation. Water absorbs infra-red radiation with wavelengths near 2.80um. Suppose this radiation is absorbed by the water and converted to heat. A 1.00 L sample of water absorbs infra-red radiation, and its temperature increases from 20C to 30C. How many photons of this radiation are used to heat the water?
My work:
q = 1000 x 4.184 x 10
q = 41840
detalE = hc/wavelenght
wavelenght = 2.8 x 10^-6 m
deltaE = [(6.626 x 10^-34)(3.0 x 10^8)]/2.8 x 10^-6
delta E = 7.09 x 10^-20 m per photon
Proportion:
1 photon = 7.099 x 10^-20m
q = 41840
Therefore, 5.894 x 10^-16
Is this correct?
Warm objects emit electromagnetic radiation in the infra-red region. Heat lamps employ this principle to generate infra-red radiation. Water absorbs infra-red radiation with wavelengths near 2.80um. Suppose this radiation is absorbed by the water and converted to heat. A 1.00 L sample of water absorbs infra-red radiation, and its temperature increases from 20C to 30C. How many photons of this radiation are used to heat the water?
My work:
q = 1000 x 4.184 x 10
q = 41840
detalE = hc/wavelenght
wavelenght = 2.8 x 10^-6 m
deltaE = [(6.626 x 10^-34)(3.0 x 10^8)]/2.8 x 10^-6
delta E = 7.09 x 10^-20 m per photon
Proportion:
1 photon = 7.099 x 10^-20m
q = 41840
Therefore, 5.894 x 10^-16
Is this correct?
Answers
Answered by
bobpursley
No, not hardly. I get about a mole of photos in my head.
What is q/energyonephoton ?
What is q/energyonephoton ?
Answered by
GenChem
q=heat absorbed by water = mass x specific heat water x delta T
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