Enormous numbers of microwave photons are needed to warm macroscopic samples of matter. a portion of soup containing 167 g of water is heated in a microwave oven from 20.0°c to 98.0°c, with radiation of wavelength 1.55 × 10−2 m. how many photons are absorbed by the water in the soup?

4.2 x 10²⁷ photons

Explanation:

Remember that the energy of a single photon is given by

     E = hc/λ  where λ is the wavelength, c speed of light, and h Planck's constant.

So to solve the problem we need to calculate the energy in joules necessary to heat the water and then divide by the enegy above for a photon.

Ephoton= 6.626 x 10 ⁻³⁴ Js x  3 x 10^8 m/s / 1.55 x 10⁻² m

= 1.28 x 10⁻²³ J/photon

ΔHs=  m x C x ΔT

T₁ = 20.0 + 273 = 293 K

T₂=  98.0 + 273 = 371 K

C= 4.184 J/ g K    ( Specific Heat of water )

ΔHs= 167 g x 4.184 j/gK x ( 371 - 293 ) K = 5.45 x 10⁴ J

# photons = 5.45 x 10⁴ J/ 1.28 x 10⁻²³ J/photon  = 4.2 x 10 ²⁷ photons

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