Question

Chemistry 121 Name Ch. 7 Challenge Problem (out of 30 points) Microwave ovens use microwave radiation to heat food. The microwaves are absorbed by moisture in the food, which is transferred to other components of the food. As the water becomes hotter, so does the food. Suppose that the microwave radiation has a wavelength of 11.2 cm. How many photons are required to heat 1 cup (8 fl. oz.) of coffee from 23°C to 60°C? Clearly state any assumptions you make in solving the problem. EL

Answer 1

Given that:

Volume of coffee = 8 fl.oz = 236.588 mL

because, 1 fl.oz = 29.5735 mL

Therefore, Mass of coffee = volume x density = (236.588 mL) x (1 g/mL) = 236.588 g

Specific heat capacity of coffee = heat capacity of water = 4.184 J/g^oC

Energy required to heat the coffee is given by:

H = mcΔT

H = (236.588 g) x (4.184 J/g^oC) x (60-23)^oC

H = 36625.7 J

Energy per photon of the microwave radiation:

E = hc/λ

Where, c = speed of light = 3 x 10⁸ m/s

E = (6.626 x 10^-34 J.s x 3 x 10⁸ m/s) / (0.112 m)

E = 1.7748 x 10^-24 J/photon

Hence No.of photons required to heat the coffee:

N = Total energy required / energy per photon

N = (36625.7 J) / (1.7748 x 10^-24 J per photon)

N = 2.064 x 10²⁸ photons ---------- (answer).

Assumptions we made to solve this problem:

• Specific heat capacity of coffee = Specific heat capacity of water = 4.184 J/g^oC
• Density of coffee = Density of water = 1 g/mL
• Energy from microwave is completely converted to heat energy of coffee (that means, energy from photon is not wasted in other manners)