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Chemistry 121 Name Ch. 7 Challenge Problem (out of 30 points) Microwave ovens use microwave radiation to heat food. The micro
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Answer #1

Given that:

Volume of coffee = 8 fl.oz = 236.588 mL

because, 1 fl.oz = 29.5735 mL

Therefore, Mass of coffee = volume x density = (236.588 mL) x (1 g/mL) = 236.588 g

Specific heat capacity of coffee = heat capacity of water = 4.184 J/goC

Energy required to heat the coffee is given by:

H = mcΔT

H = (236.588 g) x (4.184 J/goC) x (60-23)oC

H = 36625.7 J

Energy per photon of the microwave radiation:

E = hc/λ

Where, c = speed of light = 3 x 108 m/s

E = (6.626 x 10-34 J.s x 3 x 108 m/s) / (0.112 m)

E = 1.7748 x 10-24 J/photon

Hence No.of photons required to heat the coffee:

N = Total energy required / energy per photon

N = (36625.7 J) / (1.7748 x 10-24 J per photon)

N = 2.064 x 1028 photons ---------- (answer).

Assumptions we made to solve this problem:

  • Specific heat capacity of coffee = Specific heat capacity of water = 4.184 J/goC
  • Density of coffee = Density of water = 1 g/mL
  • Energy from microwave is completely converted to heat energy of coffee (that means, energy from photon is not wasted in other manners)
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