Question
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?
Answers
q=heat absorbed by water = mass x specific heat water x delta T. Assuming the density of water is 1.00 g/L, the mass of 1.0 L is 1,000 grams.
q = 1000 x 4.184 J/g*K x 10.
delta E = hc/wavelength.
Plug in h, c, and wavelength to calculate delta E. That is the energy per photon. Then set up a proportion to determine the number of photons required to reach q.
Post your work if you get stuck.
q = 1000 x 4.184 J/g*K x 10.
delta E = hc/wavelength.
Plug in h, c, and wavelength to calculate delta E. That is the energy per photon. Then set up a proportion to determine the number of photons required to reach q.
Post your work if you get stuck.
Thanks for the speedy response!
Here is 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? Thanks again for you help!
Here is 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? Thanks again for you help!