Question

the heat of fusion of ice is 6.00 kj/mol. what number of photons of wavelength = 6.42×10e-6m are needed to melt one gram of ice?

Answer 1

Given data is we have 1 gram of ice . we have to convert the value from grams to moles .

therefore number of moles of ice = 1 gram / molecular mass of H2O = 1/18 = 0.0555 moles of ice

given heat of fusion of ice is = 6.00 kj / mol .that is for 1 mole of ice heat required is = 6 kj . therefore heat required for 0.0555 moles of ice = 6 kj / mol * ( 0.0555 moles ) = 0.333 kj = 333 joules ( since 1 kj = 1000 J) .now we have energy E = 333 joules .

now from the equation E = n * h* c / wavelength

where E = energy = 333 joules

n = number of photons = ?

h = planck's constant = 6.632 * 10^-34   joule * second

c = speed of light = 3 * 10 ⁸ metre / second

wavelength = 6.42 * 10 ^-6 metre.

therefore substituting all the values in the equation we get

333 j = n *  (6.632 * 10^-34   joule * second) * ( 3 * 10 ⁸ metre / second) / (  6.42 * 10 ^-6 metre )

333 joule = n * 3.0991 * 10 ^-20 joule

therefore n = 333 / ( 3.0991 * 10 ^-20 )

n = number of photons = 1.0745 * 10 ²² photons .