Question

The heat of fusion of ice is 6.00 kJ/mol. Find the number of photons of wavelength = 6.27 10-6 m that must be absorbed to melt 3.00 g of ice.

Answer 1

Given, Heat of fusion of ice (phase transition from solid to liquid) = 6.0 kJ/ mol

Number of moles in 30. gram water = Mass / molecular mass

                                                = 6.0 g/ 18.0 g mol^-1 = (1/3) moles = 0.33333 moles

Now, amount of heat required to melt ice = molar heat of fusion x number of moles

                                                = 6 kJ mol^-1 x (1/3 moles) = 2 kJ mol

Energy associated with 1 photon of specified wavelength is given by, E = hn -----equation 1

                        Where, h = Plank’s constant   ; n = frequency of the photon (EMR)

For an EMR, c = v l

            Or, n = c/l ------ equation 2              ,[l] = wavelength

Now, l = 6.27 x 10^-6 m

            c = 3 x10⁹ ms^-1

            h = 6.626 x 10^-34 Js

From equation 2

            Frequency, v = c / l = 3.0 x 10⁸ ms^-1 / 6.27 x 10^-6 m = 0.478469 s^-1

Putting the values in equation 1,

Energy of 1 photon, E = 6.626 x 10^-34 Js x 0.478469 s^-1

= 3.17 x 10^-34 J

= 3.17 x 10^-37 kJ                                 [1 J = 10^-3 kJ]

Number of photons required to produce 2.0 kJ energy

                                    = 2.0 kJ / (3.17 x 10^-37 kJ)

                                    = 6.3 x 10³⁶ photons