A container with 0.317 L of water is placed into microwave and is then radiated with electromagnetic energy that has a wavelength of 13.3 cm. The temperature of the water then rose by 74.3 °C. 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
E = h c / lambda where
E is the energy
h is Plack's constant, h = 6.626 × 10 -34 joule·s
c is the speed of light in m/sec
lambda is the wavelength.
The total energy required = mass of water specific heat * temperature change
Divide the total energy by the energy per photon.
Answer: The number of photons absorbed by the water is \(6.59\times 10^{28}\)
Explanation:
\(E=\frac{hc}{\lambda}\)
where,
E = energy of the photon
h = Planck's constant = \(6.626\times 10^{-34}Js\)
c = speed of light = \(3.0\times 10^8m/s\)
\(\lambda\) = wavelength of photon = 13.3 cm = 0.133 m (Conversion factor: 1 m = 100 cm )
Putting values in above equation, we get:
\(E=\frac{6.626\times 10^{-34}Js\times 3.0\times 10^8m/s}{0.133m}\\\\E=1.495\times 10^{-24}J\)
To calculate the mass of water, we use the equation:
\(\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}\)
Density of water = 1 g/mL
Volume of water = 0.317 L = 317 mL (Conversion factor: 1 L = 1000 mL )
Putting values in above equation, we get:
\(1g/mL=\frac{\text{Mass of water}}{317mL}\\\\\text{Mass of water}=(1g/mL\times 317mL)=317g\)
\(q=mc\Delta T\)
where,
q = heat absorbed
m = mass of water = 317 g
c = specific heat capacity of water = 4.184 J/g°C
\(\Delta T\) = change in temperature = 74.3°C
Putting values in above equation, we get:
\(q=317g\times 4.184J/g^oC\times 74.3^oC=98546.2J\)
\(\text{Number of photons}=\frac{\text{Total amount of energy absorbed}}{\text{Energy of 1 photon}}\)
Putting values in above equation, we get:
\(\text{Number of photons}=\frac{98546.2J}{1.495\times 10^{-24}J}=6.59\times 10^{28}\)
Hence, the number of photons absorbed by the water is \(6.59\times 10^{28}\)
