Question

A microwave operates at \(250 .\) W. Assuming that the waves emerge from a point source emitter on one side of the oven, how long does it take to melt an ice cube \(2.00 \mathrm{~cm}\) on a side that is \(10.0 \mathrm{~cm}\) away from the emitter if \(10.0 \%\) of the photons that strike the cube are absorbed by it? How many photons of wavelength \(10.0 \mathrm{~cm}\) hit the ice cube per second? Assume a cube density of \(0.960 \mathrm{~g} / \mathrm{cm}^{3}\)

Short answer

Answer: The number of photons hitting the ice cube per second is approximately \(1.26 \times 10^{24}\) photons/s. However, since the energy required to melt the ice cube is 0 J, the microwave does not melt the ice cube.

Step-by-step solution

Calculate the volume of the ice cube.

We have the side length of the ice cube (\(2.00 \mathrm{~cm}\)). To find the volume, we can simply cube the side length: \(Volume = side^3\). \(Volume = (2.00 \mathrm{~cm})^3 = 8.00 \mathrm{cm}^3\)

Calculate the mass of the ice cube.

We are given the density of the ice cube, which is \(0.960 \mathrm{g} / \mathrm{cm}^3\). To find the mass of the ice cube, we can use the formula: \(mass = density \times volume\). \(Mass = 0.960 \mathrm{g/cm^3} \times 8.00 \mathrm{cm}^3 = 7.68 \mathrm{g}\)

Calculate the energy needed to melt the ice cube.

To calculate the energy needed to melt the ice cube, we need the mass of the ice, the specific heat capacity of ice (approximately \(2.1 \mathrm{J/gK}\) ), and the melting temperature of ice (\(0^\circ C = 273.15K\)). The formula for energy required is: \(Energy = mass \times specific~heat~capacity \times (T_{final} - T_{initial})\) Assuming the initial temperature of the ice cube is \(0^\circ C (273.15K)\), and the final temperature is also \(0^\circ C (273.15K)\). \(Energy = 7.68 \mathrm{g} \times 2.1 \mathrm{J/gK} \times (273.15K - 273.15K) = 0 \mathrm{J}\)

Calculate the energy received per second by the ice cube.

We are given that the microwave operates at \(250W\), which means it emits \(250 \mathrm{J}\) of energy per second. We are also given that \(10.0 \%\) of photons are absorbed by the ice cube. So, the energy received per second by the ice cube is \(0.10 \times 250 W = 25W\)

Calculate the time required to melt the ice cube.

Given the energy needed to melt the ice cube and the energy received per second, we can calculate the time required to melt the ice cube using the formula: \(Time = \frac{Energy}{Energy~received~per~second}\). Since the energy needed to melt the ice cube is \(0J\), the time required to melt the ice cube would also be \(0\).

Calculate the number of photons hitting the ice cube per second.

To find the number of photons per second, we can use the energy received per second (\(25W\)) and the energy of each photon. The energy of a photon (E) can be found using the formula: \(E = \frac{hc}{\lambda}\), where h is the Planck's constant (\(6.63 \times 10^{-34} \mathrm{J s}\)), c is the speed of light (\(3.00 \times 10^8 \mathrm{m/s}\)), and \(\lambda\) is the wavelength. Given the wavelength of the photons as \(10.0 \mathrm{cm} (0.1 \mathrm{m})\), we can calculate the energy of each photon. \(E = \frac{6.63 \times 10^{-34} \mathrm{J s} \cdot 3.00 \times 10^8 \mathrm{m/s}}{0.1 \mathrm{m}} = 1.989 \times 10^{-25} \mathrm{J}\) Now, we can calculate the number of photons per second. \(Number~of~photons~per~second = \frac{25 \mathrm{J}}{1.989 \times 10^{-25} \mathrm{J}} = 1.26 \times 10^{24}\ \mathrm{photons/s}\)