Question

Suppose that the microwave radiation has a wavelength of 12 cmcm . How many photons are required to heat 245 mLmL of coffee from 25.0 ∘C∘C to 62.0 ∘C∘C? Assume that the coffee has the same density, 0.997 g/mLg/mL , and specific heat capacity, 4.184 J/(g⋅K)J/(g⋅K) , as water over this temperature range.

Answer 1

2.281*10^28 photons

Step-by-step explanation:

Data

• Volume of coffee, V = 245 mL

• Temperature change of the coffee, ΔT = 62 - 25 = 37 °C

• Density of coffee, D = 0.997 g/mL

• Specific heat capacity of coffee, cp = 4.184 J/(g*K) = 4.184 J/(g* °C)

• wavelength of the microwave radiation, λ = 12 cm = 0.12 m

Mass of the coffee

m = D*V = 0.997 * 245 = 244.265 g

Energy (heat) absorbed by the coffee

Q = m*cp*ΔT = 244.265 * 4.184 * 37 = 37814.1761 J

Energy of the photon

E = h*c/λ

where h is the Planck constant ( 6.63 x 10^(-34) J*s) and c is the speed of light at vacuum (3*10^8 m/s)

E = 6.63 x 10^(-34)*3*10^8/0.12 = 1.6575*10^(-24) J

number of photons required = total energy required / energy per photon

n = 37814.1761/1.6575*10^(-24) = 2.281*10^28 photons