Question

g A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat. If the radiation wavelength is 12.5 cm, how many photons of this radiation would be required to heat a container with 0.250 L of water from a temperature of 20.0oC to a temperature of 99oC

Answer 1

Answer:

The total photons required = 5.19 × 10²⁸ photons

Explanation:

Given that:

the radiation wavelength λ= 12.5 cm = 0.125 m

Volume of the container = 0.250 L = 250 mL

The density of water = 1 g/mL

Density = mass /volume

Mass =  Volume ×  Density

Thus; the mass of the water =  250 mL ×  1 g/mL

the mass of the water = 250 g

the specific heat of water s = 4.18 J/g° C

the initial temperature $T_1$ = 20.0° C

the final temperature $T_2$ = 99° C

Change in temperature $\Delta T$ = (99-20)° C = 79 ° C

The heat q absorbed during the process = ms $\Delta T$

The heat q absorbed during the process = 250 g × 4.18 J/g° C × 79° C

The heat q absorbed during the process = 82555 J

The energy of a photon can be represented by the equation :

= hc/λ

where;

h = planck's constant = $6.626 \times 10^{-34} \ J.s$

c = velocity of light = $3.0 \times 10^8 \ m/s$

=  $\dfrac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{0.125}$

= $1.59024 \times 10^{-24}$ J

The total photons required = Total heat energy/ Energy of a photon

The total photons required = $\dfrac{82555 J}{1.59024 \times 10^{-24}J}$

The total photons required = 5.19 × 10²⁸ photons