Microwave ovens heat food by creating microwave electromagnetic radiation that is absorbed by water molecules in the food.
Any material that does not have water in it will not absorb the radiation and will not get hot. Metals reflect the microwaves from their surfaces and disrupt the operation of the oven.
Some Australians heat water in microwave ovens to make tea.

What is the minimum number of microwave photons with a wavelength of 3.02 mm that will have to be absorbed by a 110.00 gram sample of water to heat it from 25°C to 100°C?

(It requires about 315 joules of energy to heat 1 gram of water 75°C)

Respuesta :

Explanation:

It is known that relation between heat energy and temperature change is as follows.

             Q = [tex]mC \Delta T[/tex]

where, q = heat energy

            m = mass of the solution or solvent

     [tex]\Delta T[/tex] = change in temperature

Hence, putting the given values into the above formula as follows.

            Q = [tex]mC \Delta T[/tex]

                = [tex]110.0 g \times \frac{315 J}{75^{o}C} \times (100 - 25)^{o}C[/tex]

                = 34650 J

Now, calculate the energy of photons at wavelength 3.02 mm as follows.

             E = [tex]\frac{hc}{\lambda}[/tex]

where,     h = planks constant = [tex]6.626 \times 10^{-34} Js[/tex]

                c = velocity of light = [tex]3 \times 10^{8}[/tex] m/s

            [tex]\lambda[/tex] = wavelength = 3.02 mm = [tex]3.02 \times 10^{-3}[/tex] m   (as 1 m = 1000 mm)

Therefore, putting the given values into the above formula as follows.

              E = [tex]\frac{hc}{\lambda}[/tex]

                 = [tex]\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{3.02 \times 10^{-3} m}[/tex]

                = [tex]6.58 \times 10^{-23}[/tex] J

Now, the number of photons required of energy [tex]6.58 \times 10^{-23}[/tex] J/photon for the total energy of 34650 J as follows.

                [tex]E_{total} = n \times E[/tex]

                     n = [tex]\frac{E_{total}}{E}[/tex]

                        = [tex]\frac{34650 J}{6.58 \times 10^{-23}}[/tex]

                        = [tex]5265.95 \times 10^{23}[/tex] photons

or,                     = [tex]5.265 \times 10^{26}[/tex] photons

Thus, we can conclude that the minimum number of photons present are [tex]5.265 \times 10^{26}[/tex].

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