Question

A container with 0.389 L of water is placed in a microwave and radiated with electromagnetic energy with a wavelength of 12.9 cm. The temperature of the water rose by 78.7 °C. Calculate the number of photons that were absorbed by the water. Assume water has a density of 1.00 g ⋅ mL − 1 and a specific heat of 4.184 J ⋅ g − 1 ⋅ ° C − 1 .

Answer 1

Explanation:

The given data is as follows.

       V = 0.389 L = 0.389 kg

        m = 389 g of water       (as 1 kg = 1000 g)

   wavelength = 12.9 cm

        dT = $78.7^{o}C$

First, we will identify the heat absorbed as follows.

            Q = $m \times C \times (T_{f} - T_{i})$

Putting the given values into the above formula as follows.        

          Q = $m \times C \times (T_{f} - T_{i})$

          Q = $389 g \times 4.184 \times (78.7^{o}C)$

               = 128090.231 J

Now, as we know that

            Energy (E) = $\frac{hc}{\lambda}$

or,          $\lambda = \frac{hc}{E}$

where,     h = Planck's Constant = $6.626 \times 10^{-34}$ J s

                c = speed of particle (i.e. light) = $3 \times 10^{8}$ m/s

                E = energy per particle J/photon

Hence, calculate the energy as follows.

       E = $\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.129 m}$

(As 1 m = 100 cm)          

         = $1.5409 \times 10^{-24}$ J/photons

Hence, number of photons present will be calculated as follows.

   n photons = $\frac{Q}{E}$

                 = $\frac{128090.231 J}{1.5409 \times 10^{-24}}$

                 = $83126.89 \times 10^{24}$

                 = $8.312 \times 10^{28}$

Thus, we can conclude that number of photons released are $8.312 \times 10^{28}$.