Assume that water absorbs light of wavelength 3.00x10^-6 m with 100% efficiency. how many photons are required to heat 1.00g of water by 1.00K? the heat capacity of water is 75.2J/mol K.
all i have so far is hc/ lambda.
2007-01-16 14:16:51 · 2 answers · asked by Windy - 1 in Science & Mathematics ➔ Chemistry
So far, so good.
hc/lambda is the energy contained in a single photon, where h is Planck's constant, c is the speed of light, and lambda is your wavelength.
Find out how much energy is required to raise the temperature of the water using q = mc delta T. In this case, q is the energy, m is the mass of the water (1.00 g), c is the specific heat of water (4.186 J/g-K), and delta T is the change in temperature (1 K). This gives a result of (1)(4.186)(1) = 4.186 J.
Divide the energy per photon into this value and you are done.
Note that the c (speed of light) in your first equation is different from the c (specific heat) in your second equation. Just one of the annoyances of physics notation :)
2007-01-16 14:31:01 · answer #1 · answered by Stephen S 3 · 0⤊ 0⤋
Well, 1 gram of water is 1/18 mol. So you'll need 75.2/18 = 4.18 joules. The energy of a photon is hc/L where h= Planck's constant 6.67E-34, c is the speed of light 3E8 and L=wavelength, in this case 3E-6, so the number of photons needed is 4.18 times 3E-6/6.67E-34/3E8=6.27E19.
2007-01-16 22:30:33 · answer #2 · answered by zee_prime 6 · 0⤊ 0⤋