numberphotons*Energy/photon=masscoffie*c*changetemp
= 295*.997*4.18*(62-25) joules
Now the energy/photon= h lambda/speedlight=6.6e-34 m2 kg / s *.12m/3e8m/s
= 6.6e-34*.12/3e8 joules
so finally, number jphotons= (295*.997*4.18*(62-25))/(6.6e-34*.12/3e8)
= 1.72e47 check that.
Suppose that the microwave radiation has a wavelength of 12 cm . How many photons are required to heat 295 mL of coffee from 25.0 ∘C to 62.0 ∘C? Assume that the coffee has the same density, 0.997 g/mL , and specific heat capacity, 4.184 J/(g⋅K) , as water over this temperature range.
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