Question

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.

Answer 1

step 1: energy required to heat coffee E = m Cp dT E = energy to heat coffee m = mass coffee = 225 mL x (0.997 g / mL) = 224g Cp = heat capacity of coffee = 4.184 J / gK dT = change in temp of coffee = 62.0 - 25.0 C = 37.0 C E = (224 g) x (4.184 J / gK) x (37.0 C) = 3.46x10^4 J step2: find energy of a single photon of the radiation E = hc / λ E = energy of the photon h = planck's constant = 6.626x10^-34 J s c = speed of light = 3.00x10^8 m/s λ = wavelength = 11.2 cm = 11.2 cm x (1m / 100 cm) = 0.112 m E = (6.626x10^-34 J s) x (3.00x10^8 m/s) / (0.112 m) = 1.77x10^-16 J step3: Number of photons 3.46x10^4 J x ( 1 photon / 1.77x10^-16 J) = 1.95x10^20 photons

Answer 2

Answer:

$n = 1.96 \times 10^{28}$ photons

Explanation:

Mass of the coffee is given as

$mass = density\times volume$

here we know that

density = 0.997 g/ml

volume = 225 ml

now mass of coffee will be

$m = 0.997 \times 225 = 224.325 g$

now heat required to raise the temperature of coffee will be

$Q = ms\Delta T$

$Q = 224.325 \times 4.184 \times (62 - 25)$

$Q = 34727.3 J$

now we know that energy of one photon is

$E = \frac{hc}{\lambda}$

$E = \frac{(6.6 \times 10^{-34}) ( 3 \times 10^8)}{0.112}$

$E = 1.77 \times 10^{-24} J$

now it requires "n" photons to complete the energy

$n\times (1.77 \times 10^{-24}) = 34727.3$

$n = 1.96 \times 10^{28}$