Question

A container with 0.389 L of water is placed in a microwave and is then radiated with electromagnetic energy that has a wavelength of 11.3 cm. The temperature of the water then rose by 67.7C. Calculate the number of photons that were absorbed by the water. Assume water has a density of 1.00 g*mL^-1 and its specific heat is 4.184 J*g^-1*C^-1

Answer 1

VOlume of water = 0.389 L= 389 mL

Density of water = 1.00g/mL

Mass of water = VOlume * density = 389 mL * 1.00g/mL = 389 g

THus, HEat absorbed by water = mass * HEat capacity * change in temperature

= 389 g *(4.184J/g.^oC) * 67.7 ^oC = 110186.90J

Wavelength of energy radiated = 11.3 cm = 0.113 m

Thus, energy = hc/wavelength = (6.626*10^-34 J.s)* (3*10⁸m.s^-1)/ 0.113 m = 1.759 *10^-24J

Number of photons = Heat absorbed/ Energy radiated

= 110186.90J /(1.759 *10^-24J) = 6.26*10²⁸photons