Tricky Chemistry Problem help?

Question

I'm totally stuck with this problem...
A certain microwave oven delivers 750 watts (joule/s) of power to a coffee cup containing 50 g of water at 25 degrees C. If the wavelength of microwaves in the oven is 9.75cm, how long does it take, and how many photons must be absorbed to make the water boil? The specific heat capacity of water is 4.18J/Cg and assume only the water absobs the energy of the microwaves. Thanks so much!

Answer 1

Who cares about photons if the microwave does not operate at a proper frequency?

1. The wave length of the microwave is only of one significance that is if it will resonate water molecules to heat it up. Most photons, that do not resonate with water molecule, will just pass through the water without heating it up.
Since you have a wavelength (l) of 9.75 cm or .0975m.
if c-is the speed of light and f is the frequency then
f=c/l=(300,000,000m/s)/(.0975m )=
f=3,076,923,077Hz
f= 3.077GHz
The resonant frequency for water is 2.5GHz. (We are close but there could be an efficiency problem :-) It should be about 12cm.)

2. It says 750 watts of power. What is the efficience of electromagnetic energy conversion? We can assume nearly 100%.

t=E/P
t=time
E - energy
P - power

E=mp(T2-T1)
m-mass of water
p - specific heat
T- temperature

so
t=mp(T2-T1)/P=
t=50 x 4.14(100-25)/750=
t=20.9 seconds

Ep=hf
Ep= photon energy
h - Plank's constant ( 6.626 x 10^-34 Joules Sec)

If N - Number of photons
Then
N= E/Ep
N=mp(T2-T1)/hf
N=50x4.18(100-25)/( 6.626 x 10^-34 3.077 x 10 9)=
N=7.688x10^27 photons

For more info see the reference

Answer 2

It's more of a thermal physics than a chemistry problem. The water must be heated by 75 C to boil (we assume sea level). You can calculate the total energy required to do this from the amount of water and its heat capacity. Dividing this by 750 watts will give you the time.

The photon part of the problem requires some simple quantum mechanics, in particular the formula E = h / lambda, where E is the energy of a photon, h is Planck's constant, and lambda is the wavelength of the radiation (given -- incorrectly -- it should be 14.3 cm.). Look up Planck's constant, do the obvious division, and get the energy per photon. You know the total energy, so one division will give you the number of photons required to carry that much energy. It is a BIG number.

Answer 3

One correction to the above. The equation is
E = hc/lambda
where E=energy, h is Planck's constant, c is the speed of light in vacuum and lambda is the wavelength in meters.