Photons Required To Heat Coffee In A Microwave Oven

Strings (SiPjAjk) = S7P3A32     Base Sequence = 12735     String Sequence = 12735 - 3 - 32

Expressions Of Pj Problems
Photons Required To Heat Coffee In A Micrwave Oven
Math Microwave ovens use microwave radiation to heat food. Moisture in the food absorbs microwaves. Food becomes hotter as moisture in food becomes hotter.

How many photons must a microwave radiation with a wavelength of 11.2 cm produce, in order to heat 200 mL of coffee in a microwave oven, from 23o C to 60o C?

The strings:

S7P3A32 (Force - Push).

The math:
Pj Problem of Interest is of type force (push). Energy is the capacity to do work which is actualize by force. Light energy can do a pull or push work. The push work is more common. In this problems the photons are being pushed out. Hence the Pj Problem of Interest here is of type force-push. Energy in one photon = E = hν-----------(1).
Where E is the energy of one photon; h is Planck's constant (0.66252 x 10-33 Js) and ν is the frequency of the microwave.
ν = (velocity of light)/(wavelength of light)
= (3 x 108)/(11.2 x 10-2) = 26.79 x 108 Hz Hertz cycles per second).
So, Energy of one photon of microwave with wavelength 11.2 cm is:
= hν = (0.66252 x 10-33)(26.79 x 108) = 17.75 x 10-25 J.

Specific heat of water = 4.184 J.
Density of water = 1gm/mL
Assuming coffee is dilute aqueous solution
So, specific heat of coffee = specific heat of water.
So, Energy required to heat 200 mL of coffee from 23o C to 60o C
= 4.184 x 200 x (60 -23) = 4.184 x 200 x 37 = 30.961 x 103.

So, number of photons required to heat coffee in the microwave oven from 23o to 60o when the wavelength of the microwave radiation is 11.2 cm is:
(30.961 x 103)/(17.75 x 10-25) = 1.7 x 1028 The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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