heat (energy) needed ... 295 * (62.0 - 25.0) * 0.997 * 4.184 Joules
energy per photon = h * c / 12 = 6.63E-34 * 3.00E10 / 12 Joules
divide the heat needed by the energy per photon
Suppose that the microwave radiation has a wavelength of 12 cm . How many photons are required to heat 295 mL of coffee from 25.0 ∘C to 62.0 ∘C? Assume that the coffee has the same density, 0.997 g/mL , and specific heat capacity, 4.184 J/(g⋅K) , as water over this temperature range. the answer WAS NOT 1.72e47
2 answers
heat needed = q = mass x specific heat x (Tfinal-Tinitial)
q = 295 mL x 0.997 g/mL x 4.184 x (62-25) = approx 46,000 J but you need a better answer and not this estimate.
How much energy in joules can you get from 12 cm light.
That is E in joules/photon = hc/wavelength = 6.626E-34 x 3E8/0.12 = 1.66E-24 J.
Then 1.66E-24 J/photon x #photons = approx 46,000 J
Solve for # photons.
q = 295 mL x 0.997 g/mL x 4.184 x (62-25) = approx 46,000 J but you need a better answer and not this estimate.
How much energy in joules can you get from 12 cm light.
That is E in joules/photon = hc/wavelength = 6.626E-34 x 3E8/0.12 = 1.66E-24 J.
Then 1.66E-24 J/photon x #photons = approx 46,000 J
Solve for # photons.