Question

Number of photons used from microwave to heat water?

A microwave oven operating at `1.22* 10^8 nm` is used to heat `145 mL` of water (roughly the volume of a teacup) from `21.0°C` to `100.0°C.` Calculate the number of photons needed if `93.9` percent of microwave energy is converted to the thermal energy of water.

Answer 1

`3.13 * 10^(28)"photons"`

Explanation:

!!LONG ANSWER !!

For starters, you need to know the mass of water present in your sample. Since no mention of what to use for the density of water, you can assume it to be equal to `"1 g mL"^(-1)`.

This means that your sample contains

`145 color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "145 g"`

Now, you know that the amount of heat needed to heat the water from `21.0^@"C"` to liquid water at `100.0^@"C"` can be calculated using the equation

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `q` is the heat lost or gained by the water
• `m` is the mass of the water
• `c` is the specific heat of water,, equal to `"4.18 J g"^(-1)""^@"C"^(-1)`
• `DeltaT` is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

In your case, you have

`DeltaT = 100.0^@"C" - 21.0^@"C" = 79.0^@"C"`

which means that the heat needed for this sample is

`q = 145 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 79.0color(red)(cancel(color(black)(""^@"C")))`

`q = "47,882 J"`

Next, convert the wavelength of the photons to frequency by using the fact that wavelength and frequency have an inverse relationship described by the equation

`color(blue)(ul(color(black)(nu * lamda = c)))`

Here `c` is the speed of light in a vacuum, usually given as `3 * 10^8"m s"^(-1)`.

You can say that a photon of wavelength

`1.22 * 10^8 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9color(red)(cancel(color(black)("nm")))) = "0.122 m"`

will have a frequency of

`nu = c/(lamda)`

`nu = (3 * 10^8 color(red)(cancel(color(black)("m"))) "s"^(-1))/(0.122color(red)(cancel(color(black)("m")))) = 2.459 * 10^9` `"s"^(-1)`

Next, calculate the energy of a single photon by using the Planck - Einstein equation

`color(blue)(ul(color(black)(E = h * nu)))`

Here

• `E` is the energy of the photon
• `h` is Planck's constant, equal to `6.626 * 10^(-34)"J s"`
• `nu` is the frequency of the photon

In your case, you will have

`E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 2.459 * 10^9color(red)(cancel(color(black)("s"^(-1))))`

`E = 1.629 * 10^(-24)` `"J"`

Now, calculate the number of photons needed to heat the water assuming that you have an `100%` efficiency

`"47,882" color(red)(cancel(color(black)("J"))) * "1 photon"/(1.629 * 10^(-24)color(red)(cancel(color(black)("J")))) = 2.939 * 10^(28)` `"photons"`

So if all the photons supplied by the microwave would be used to heat the water, you would need `2.939 * 10^(28)` photons to heat your sample.

However, you know that only `93.9%` of the energy supplied by the microwave is being converted to heat, which means that out of `1000` photons supplied by the microwave, only `939` photons are used to heat the water.

You can thus say that the total number of photons that must be supplied by the microwave in order for `2.939 * 10^(28)` photons to heat the water is equal to

`2.939 * 10^(28) color(red)(cancel(color(black)("photons needed"))) * "1000 photons emitted"/(939 color(red)(cancel(color(black)("photons used"))))`

`=color(darkgreen)(ul(color(black)(3.13 * 10^(28)color(white)(.)"photons")))`

The answer is rounded to three sig figs.