Question

Suppose that the microwave radiation has a wavelength of 10.8 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. Express the number of photons numerically.

Answer 1

The microwave radiation has a wavelength of 10.8 cm = 0.108 m

So, the energy of each photon = E = h , where is the frequency and h = planck's constant. = c/λ, where λ = wavelength.

So, = c/λ = (3.00 10⁸)/0.108 = 2.78 10⁹ Hz

h = 6.626 10^-34 J/Hz

E = h = 2.78 10⁹ 6.626 10^-34 = 1.84 10^-24 J

The total energy required is
w = m C_p ∆T

m = Density Volume = 0.997 295 = 294.12 g
C_p = 4.184 J/g.K
∆T = 62 - 25 = 37^oC = 37 K

w = 294.12 4.184 37 = 45532.13 J

No. of Photons = w/E = 45532.13 / (1.84 10^-24) = 2.47 10²⁸ photons