Microwave ovens use microwave radiation to heat food. the microwaves are absorbed by the water molecules in the food, which is transferred to other components of the food. as the water becomes hotter, so does the food. part a suppose that the microwave radiation has a wavelength of 11.2 cm . how many photons are required to heat 225 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.

The number of photons that are required to heat 225 ml of coffee from 25.0 °C to 62 °c

n = 1.962 × 10^(28) photons

We are given;

Wavelength; λ = 11.2 cm = 0.112 m

density; ρ = 0.997 g/ml

volume; V = 225 ml

Specific heat capacity; c = 4.184 J/g.k

Change in temperature;

Δt = (273 + 62) - (273 + 25) K = 37 K

Formula for mass given density and volume is;

m = Vρ

Thus; mass of coffee;

m = 225 × 0.997

m = 224.325 g

Now, formula for Quantity of heat required to raise the temperature is;

Q = mcΔt

Thus;

Q = 224.325 × 4.184 × 37

Q = 34,727.3046 J

Now, we know that formula for energy of photon is; E = hc/λ

Where;

h is Planck's constant = 6.6 × 10^(-34) J.s

c is speed of light = 3 × 10^(8) m/s

λ is wavelength.

Thus;

E = [6.6 × 10^(-34) × 3 × 10^(8)]/0.112

E = 1.77 × 10^(-24) J

To get the number of photons required, we will use the formula;

n = Q/E

Thus;

n = 34,727.3046/(1.77 × 10^(-24))

n = 1.962 × 10^(28) photons

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photons

Explanation:

Mass of the coffee is given as

here we know that

density = 0.997 g/ml

volume = 225 ml

now mass of coffee will be

now heat required to raise the temperature of coffee will be

now we know that energy of one photon is

now it requires "n" photons to complete the energy

this is a good question! i'll do some research and come back when i have the answer!