Question

Suppose that the microwave radiation has a wavelength of 12.4 cm. How many photons are required to heat 285 mL of coffee from 25.0 ^circ rm C to 62.0 ^circ rm C?Assume that the coffee has the same density, 0.997 g/mL, and specific heat capacity, 4.184 rm J/(gcdot K), as water over this temperature range.

Answer 1

285mL*(0.997g/mL)= 284.15g Coffee
q=mcΔT

c=heat capacity (4.184 J/g*C)

m=mass (284.15g)

ΔT= Change in temp (62.0-25.0= 37.0 C)

q=(284.15g)(4.184 J/g*C)(37.0 C)= 43987.919 J--> this is the necessary energy

h = planck constant (6.626*10^-34 Js)
c =speed of light(3*10⁸ m/s)
w = wavelength (m) =12.4cm*(1m/100cm)= 12.4*10^-2m

This represents the energy of ONE photon with a wavelength of 12.4cm, so you need to know how many equal 43987.919J