Question

A 477 g portion of soup is heated in a microwave oven from 25°C to 90°C, using radiation with a wavelength of 1.55 × 10⁻² m. Assuming the heat capacity of the soup is the same as water (4.18 J/°C･g), how many photons are absorbed by the soup?

Answer 1

To solve this problem we will use the heat transfer equations, to determine the amount of heat added to the body. Subsequently, through the energy ratio given by Plank, we will calculate the energy of each of the photons. The relationship between total energy and unit energy will allow us to determine the number of photons

The mass of water in the soup is 477g

The change in temperate is

$\Delta T = (90+273K)-(25+273K) = 65K$

Use the following equation to calculate the heat required to raise the temperature:

$q = mc\Delta T$

Here,

m = Mass

c = Specific Heat

$q = (477)(4.184)(65)$

$q = 129724.92J$

The wavelength of the ration used for heating is $1.55*10^{-2}m$

The number of photons required is the rate between the total energy and the energy of each proton, then

$\text{Number of photons} = \frac{\text{Total Energy}}{\text{Energy of one Photon}}$

This energy of the photon is given by the Planck's equation which say:

$E = \frac{hc}{\lambda}$

Here,

h = Plank's Constant

c = Velocity of light

$\lambda =$ Wavelength

Replacing,

$E = \frac{(6.626*10^{-34})(3*10^8)}{1.55*10^{-2}}$

$E = 1.28*10^{-23}J$

Now replacing we have,

$\text{Number of photons} = \frac{82240.7}{1.28*10^{-23}}$

$\text{Number of photons} = 6.41*10^{27}$

Therefore the number of photons required for heating is $6.41*10^{27}$