Question

On its high power setting, a microwave oven produces 900 W of 2560 MHz microwaves. (a) How many photons per second is this? (b) How many photons are required to increase the temperature of a 0.500-kg mass of pasta by $ 45.0 \textrm{ C}^\circ $ ,
assuming a specific heat of $0.900 \textrm{ kcal/(kg}\cdot \textrm{C}^\circ\textrm{)}$ ? Neglect all other heat transfer. (c) How long must the microwave operator wait for their pasta to be ready?

- $ 5.31 \times 10^{26} \textrm{ photons/s} $
- $ 4.99 \times 10^{28} \textrm{ photons} $
- $ 94.2 \textrm{ s} $

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This is College Physics Answers with Shaun Dychko. A microwave oven with a power of 900 watts and a frequency of 2560 megahertz is heating some pasta. The question in part (a) is how many photons per second are emitted by the microwave? Part (b) says how many total number of photons are needed to heat half a kilogram of pasta 45 Celcius degrees and part (c) is how long will this take? OK. So part (a); number of photons per second. Now the energy of a single photon which is what the subscript

*p*there is meant to denote is Planck's constant times the frequency of the photon. The total power of the microwave is the total energy of all the photons divided by time. And the total energy of all the photons is the number of photons multiplied by the energy of a single photon. And so we have here,*n*over*t*which is the number of photons per time and we'll solve for that by dividing both sides by the energy of a single photon. And so the number of photons per time is power divided by the energy for a single photon and the energy for a single photon is Planck's constant times frequency so we substitute*hf*there. So*n*over*t*then is 900 watts— power of microwave— divided by Planck's constant times 2560 times 10 to the 6 hertz which is 5.31 times 10 to the 26 and we know this is photons per second because that's what the units work out to in our equation here. So this watts can be written as joules per second, if you prefer and these joules cancel, this seconds cancels with the seconds that are in the denominator of the hertz unit because hertz is an abbreviation for 1 over seconds and so these seconds cancel leaving us with reciprocal seconds and that is photons per second. Then part (b) says what is the total number of photons needed to heat the pasta? So if you look back to chapter 14, equation 3, the energy to heat something up is equal to its mass times its specific heat multiplied by the change in temperature and we are told what this specific heat is of this pasta and we are told its mass. And the total energy can also be written as the number of photons times the energy per photon and the energy per photon is*hf*and so we can equate these two things which we do here and we'll solve for*n*, the number of photons. So divide by Planck's constant times frequency and we end up with the number of photons is mass times specific heat times change in temperature divided by Planck's constant times frequency. So that's 0.500 kilograms of pasta times the specific heat of 0.900 kilocalories per kilogram per Celsius degree and kilocalories is not an*mks*unit— meters, kilograms, seconds— so we need to convert it so that it matches with all the other units we have here. And so we'll multiply by 4186 joules per kilocalorie and now we have joules per kilogram per Celsius degree. And we times that by 45 Celsius degrees— change in temperature— divided by Planck's constant times the frequency giving us 4.99 times 10 to the 28 photons are needed to heat up the pasta. And the length of time this will take can be figured out using our power formula which is total energy divided by time and total energy is the amount of energy needed to heat it up— mass times specific heat times change in temperature— and we'll solve for time by multiplying both sides by*t*divided by power. And we have time then is*mcΔT*over*P*. So we have 0.500 kilograms times this specific heat converted into joules and then 45 Celsius degrees divided by 900 watts giving a time of 94.2 seconds.