Problem 73 - Chemistry: Principles and Reactions

Question

Microwave ovens convert radiation to energy. A microwave oven uses radiation with a wavelength of \(12.5 \mathrm{~cm}\). Assuming that all the energy from the radiation is converted to heat without loss, how many moles of photons are required to raise the temperature of a cup of water \((350.0 \mathrm{~g}\), specific heat \(=4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) ) from \(23.0^{\circ} \mathrm{C}\) to \(99.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Answer: 1. Calculate the energy required to heat the water: Q = (350 g)(4.18 J/g°C)(76°C) = 111124 J 2. Calculate the energy of one photon: E = (6.626 × 10^-34 J·s)(3.0 × 10^8 m/s) / 0.125 m = 1.59 × 10^-24 J 3. Calculate the number of photons required: n = 111124 J / 1.59 × 10^-24 J = 6.99 × 10^26 photons 4. Convert the number of photons to moles of photons: Moles of photons = 6.99 × 10^26 photons / 6.022 × 10^23 photons/mole = 1161.16 moles Therefore, 1161.16 moles of photons are required to heat the water from 23.0°C to 99.0°C.

Step 1: Calculate Energy Required to Heat the Water

First, calculate the energy (Q) required to heat the water from its initial temperature (23.0°C) to its final temperature (99.0°C) using the formula: \(Q = mcΔT\) where m is the mass of water (350 g), c is the specific heat of water (4.18 J/g°C), and ΔT is the change in temperature (99.0°C - 23.0°C). \(Q = (350 \mathrm{ ~g}) (4.18 \mathrm{ ~J}/\mathrm{g }^{\circ} \mathrm{C})(99.0^{\circ} \mathrm{C} - 23.0^{\circ} \mathrm{C})\) Calculate the energy required.

Step 2: Calculate Energy of One Photon

Next, calculate the energy (E) of one photon using Planck's equation: \(E = \dfrac{hc}{λ}\) where h is Planck's constant (\(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\)), c is the speed of light (\(3.0 \times 10^8 \mathrm{ ~m/s}\)), and λ is the wavelength of radiation given in meters (12.5 cm = 0.125 m). \(E = \dfrac{(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}) (3.0 \times 10^8 \mathrm{ ~m/s})}{0.125 \mathrm{ ~m}}\) Calculate the energy of one photon.

Step 3: Calculate Number of Photons Required

Now find the number of photons (n) required to provide the energy found in step 1, using the energy of one photon calculated in step 2: \(n = \dfrac{Q}{E}\) Calculate the number of photons required.

Step 4: Convert Number of Photons to Moles

Finally, convert the number of photons to moles of photons using Avogadro's number (\(6.022 \times 10^{23} \hspace{1mm} \mathrm{photons/mole}\)): Moles of photons = \(\dfrac{\mathrm{Number \hspace{1mm} of \hspace{1mm} Photons}}{6.022 \times 10^{23} \hspace{1mm} \mathrm{photons/mole}}\) Calculate the number of moles of photons required to heat the water.

Key Concepts

Photon Energy Calculation

To determine the amount of energy carried by a single photon, we use the photon energy calculation. This involves Planck's equation, which provides a way to connect the energy of a photon with its frequency or wavelength. The energy of a photon is given by the equation:\[ E = \frac{hc}{\lambda} \]Here:

• \( E \) is the energy of the photon.
• \( h \) is Planck's constant, valued at \( 6.626 \times 10^{-34} \ \text{J} \cdot \text{s} \).
• \( c \) is the speed of light, \( 3.0 \times 10^8 \ \text{m/s} \).
• \( \lambda \) is the wavelength of the microwave radiation, which in this problem needs converting to meters (12.5 cm = 0.125 m).

By substituting these values into the equation, you can calculate the energy for one photon of the microwave radiation used in this exercise. This step is crucial because it tells us how much energy each photon contributes to the heating process.

Specific Heat Capacity

Specific heat capacity is a material's ability to store heat energy for a given mass. It indicates how much heat is required to raise the temperature of one gram of a substance by one degree Celsius (°C).

For water, a specific heat capacity of 4.18 J/g°C means that heating one gram of water by one degree Celsius requires 4.18 Joules of energy. Specific heat is a critical factor here because it determines the total energy needed to reach a desired temperature increase.The calculation for the heat energy needed (\[ Q = mc\Delta T \]is straightforward, utilizing:

• \( m \), which is the mass of water in grams (350 g in this case).
• \( c \), the specific heat of water (4.18 J/g°C).
• \( \Delta T \), which represents the change in temperature (in this exercise, from 23°C to 99°C).

This concept helps represent the total amount of energy required to convert into heat which the photons must supply.

Temperature Change

Understanding temperature change is integral in problems dealing with energy, especially when heating a substance. Temperature change is simply the difference between the initial and final temperatures. In this exercise:

• The initial temperature of the water is 23°C.
• The final temperature is 99°C.
• Thus, the change in temperature, \( \Delta T \), is \( 99\degree C - 23\degree C = 76\degree C \).

This increase represents how much the water is heated.

This change converts directly into the Q (heat energy) calculation using the specific heat capacity equation. The calculation explains how much energy the microwave photons need to supply to achieve this temperature rise.

Planck's Equation

Planck's equation is a fundamental principle in quantum mechanics. It illustrates the quantization of energy in electromagnetic radiation. The equation is:\[ E = h u \]However, for finding photon energy given a wavelength, it is rearranged to:\[ E = \frac{hc}{\lambda} \]This equation helps relate the quantum perspective of energy (E) with physical properties like wavelength (\( \lambda \)) or frequency (\( u \)).

• \( h \), Planck's constant, captures how the energy quanta stand distinct from classical continuous energy spread.
• \( c \), speed of light, connects the wave propagation with energy.
• \( \lambda \), wavelength, shows that longer wavelengths carry less energy per quantum.

By calculating energy this way, Planck's equation demonstrates the discrete packets of energy (photons) that contribute directly to the energy conversion in examples like the microwave radiation in this exercise.