Photons Required To Heat Coffee In A Microwave Oven

**Strings (S _{i}P_{j}A_{jk}) = S_{7}P_{3}A_{32} Base Sequence = 12735 String Sequence = 12735 - 3 - 32 **

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Photons Required To Heat Coffee In A Micrwave Oven

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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 23^{o} C to 60^{o} C?

**The strings**:

S_{7}P_{3}A_{32} (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 10^{8})/(11.2 x 10^{-2}) = 26.79 x 10^{8} 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 10^{8}) = 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 23^{o} C to 60^{o} C

= 4.184 x 200 x (60 -23) = 4.184 x 200 x 37 = 30.961 x 10^{3}.

So, number of photons required to heat coffee in the microwave oven from 23^{o} to 60^{o} when the wavelength of the microwave radiation is 11.2 cm is:

(30.961 x 10^{3})/(17.75 x 10^{-25}) = 1.7 x 10^{28}

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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