Question

Suppose that the microwave radiation has a wavelength of 12.4 cm. How many photons are required to heat 305 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.

Answer 1

Density of coffee = 0.997 g/mL

Volume of coffee = 305 mL

Mass of coffee = Density Volume = 305 0.997 = 304.085 g Coffee

q = mC_pT

C_p = heat capacity = 4.184 J/(g.K)
m = mass = 304.085 g
T = change in temperature = 62-25 = 37^oC

q = 304.085 4.184 37 = 47074.79 J

To raise the temperature of the coffee, you need that many joules to heat it up. You now have to find the energy of the photon at the microwave wavelength.

E = hc/

h = planck constant = 6.626 10^-34 Js
c = speed of light = 3 10⁸ m/s
= wavelength = 12.4 cm = 0.124 m

E = (6.626 10^-34)(3 10⁸)/0.124 = 1.603 10^-24

The energy of a photon with a wavelength of 12.4 cm is 1.603 10^-24 J. You want to know how many photons would make 34727.30 J, so:

47074.79 J / (1.603 10^-24/photon) = 2.94 10²⁸ photons