Question

Suppose that the microwave radiation has a wavelength of 12 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

1. Determine the change in temperature.
dT = 62 - 25
dT = 37 c (or K)

2. Determine the mass of the water
M = v * density
M = 305 ml * 0.997 g/mL
M = 304.085 g

3. Determine the energy needed to change the temperature.
E = dT * m * specific heat constant
E = 37 K * 304.085 g * 4.184 J/(gK)
E = 47074.79 J

4. Determine the energy in a photo of electromagnetic radiation a 12cm length. h is Planks constant, c is the speed of light, v is the wavelength.

E(photon) = h * c / v
E(photon) = 6.63x10^-34 Js * 3x10^8 m/s / 0.12 m
E(photon) = 1.657x10^-24 J

5. Divide the energy required by the energy of a photon.

n = E / E(photon)
n = 47074.79 J / 1.657x10^-24 J
n = 2.84 x10^28