Question

Part A Suppose that the microwave radiation has a wavelength of 11.6 cm . How many photons are required to heat 275 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

Mass of coffee ,m = volume x density

                           = 275 mL x 0.997 (g/mL)

                           = 274.2 g

Heat required to heat the coffee , Q = mcdt

Where

m = mass of coffee = 274.2 g

c = specific heat capacity = 4.184 J/(g⋅K)

dt = change in temperature = final - initial

    = 62.0 - 25.0 ^oC

    = 37.0 ^oC

Plug the values we get Q = 42448 J

This much amount of energy is comming from radiation

We know that Energy, E = nhc / λ

Where

n = number of photons = ?

h = plank's constant = 6.625x 10^-34 Js

c = speed of light = 3x 10⁸ m/s

λ = wave length = 11.6 cm = 11.6x10^-2 m

E = energy = 42448 J

Plug the values we get

n = (E λ) / (hc)

= 2.48x10²⁸ photons   

Therefore the number of photons are 2.48x10²⁸ photons