Problem 120 - Chemistry The Molecular Science

Question

Microwave ovens, commonly used to heat water in bevcrages and foods, cmit radiation with a wavelength of \(12.2 \mathrm{~cm}\) (a) Calculate the amount (moles) of photons of this microwave radiation required to raise the temperature of \(230.0 \mathrm{~g}\) water (such as in a cup of coffee, which is mainly water) from \(24.0^{\circ} \mathrm{C}\) to \(55.0^{\circ} \mathrm{C}\) (b) As noted in Chapter \(4,\) the watt. \(W\), is a unit of power: \(1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\). If the microwave oven is rated at \(800 \mathrm{~W}\) calculate the time needed to heat the water in part (a). Assume that all the energy is delivered to the water.

Step-by-Step Solution

Answer

(a) 30.4 moles of photons; (b) 37.25 seconds.

Step 1: Calculate Energy Required to Heat Water

First, we need to find out how much energy is required to heat the water from 24.0°C to 55.0°C. This can be calculated using the formula for heat energy: \( q = m \cdot c \cdot \Delta T \) where \( q \) is the heat energy in joules, \( m \) is the mass of the water in grams (230.0 g), \( c \) is the specific heat capacity of water (4.18 J/g°C), and \( \Delta T \) is the change in temperature (55.0°C - 24.0°C = 31.0°C). Substituting these values, we get \[ q = 230.0 \text{ g} \times 4.18 \text{ J/g°C} \times 31.0 \text{ °C} = 29798.6 \text{ J} \].

Step 2: Calculate Energy of One Photon

The energy of a photon can be calculated using the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \(6.626 \times 10^{-34} \text{ J·s}\), \( c \) is the speed of light \(3.00 \times 10^{8} \text{ m/s}\), and \( \lambda \) is the wavelength of the radiation in meters. The microwave wavelength is given as 12.2 cm, which is 0.122 m. Substituting these values gives \[ E = \frac{6.626 \times 10^{-34} \text{ J·s} \times 3.00 \times 10^{8} \text{ m/s}}{0.122 \text{ m}} = 1.63 \times 10^{-24} \text{ J/photon} \].

Step 3: Calculate Moles of Photons Required

To find the moles of photons required, we divide the total energy needed by the energy per photon: \[ \text{number of photons} = \frac{29798.6 \text{ J}}{1.63 \times 10^{-24} \text{ J/photon}} \approx 1.83 \times 10^{28} \text{ photons} \]. The moles of photons can be found by dividing by Avogadro's number \(6.022 \times 10^{23} \text{ photons/mol}\): \[ \text{moles of photons} = \frac{1.83 \times 10^{28}}{6.022 \times 10^{23}} \approx 30.4 \text{ moles} \].

Step 4: Calculate Time to Heat the Water

Using the microwave oven's power rating, 800 W (which is 800 J/s), we can calculate the time required to supply 29798.6 J of energy using the formula \( \text{time} = \frac{q}{P} \), where \( q \) is the energy required, and \( P \) is the power of the microwave. Substituting the values, we have \[ \text{time} = \frac{29798.6 \text{ J}}{800 \text{ J/s}} = 37.25 \text{ seconds} \].

Key Concepts

Photon Energy Calculation

When dealing with microwave radiation, each photon carries a specific amount of energy. This energy can be determined using the formula \( E = \frac{hc}{\lambda} \). Here, \( E \) is the energy of a single photon, \( h \) is Planck's constant \(6.626 \times 10^{-34} \text{ J·s}\), \( c \) is the speed of light \(3.00 \times 10^{8} \text{ m/s}\), and \( \lambda \) is the wavelength of the radiation converted to meters. In our example, the microwave radiation has a wavelength of 12.2 cm, equivalent to 0.122 meters.
By inserting these values into the formula, we can find the energy per photon:

• \( E = \frac{6.626 \times 10^{-34} \text{ J·s} \times 3.00 \times 10^{8} \text{ m/s}}{0.122 \text{ m}} = 1.63 \times 10^{-24} \text{ J/photon} \)

This calculation is crucial because it allows us to determine how many photons are needed for a particular energy requirement.

Specific Heat Capacity

Specific heat capacity is an important concept when it comes to heating substances. It is defined as the amount of heat required to change the temperature of one gram of a substance by one degree Celsius. For water, the specific heat capacity is a well-known value, \(4.18 \text{ J/g°C}\).
This value explains why water requires a significant amount of energy to change its temperature. It means that for every gram of water, you need 4.18 joules of energy to raise its temperature by 1°C.
So, when you want to heat \(230.0 \text{ g}\) of water from 24.0°C to 55.0°C, you just utilize the formula:

• \( q = m \times c \times \Delta T \)
• \( q = 230.0 \text{ g} \times 4.18 \text{ J/g°C} \times 31.0 \text{ °C} = 29798.6 \text{ J}\)

Understanding specific heat capacity helps us calculate exactly how much energy is needed to achieve the desired temperature change.

Power and Energy Units

Energy and power are closely related but distinct concepts in physics. Power is the rate at which energy is transferred or converted. It is measured in watts (W), where 1 W is equal to 1 joule per second (J/s).
In the context of a microwave heating water, an 800 W microwave tells us that it provides 800 joules of energy every second. This rate is useful when calculating how long it will take to deliver a certain amount of energy required for heating.
For instance, if you know the energy required to heat the water is 29798.6 J, and your microwave provides 800 J/s, you can determine the time it will take:

• \( \text{time} = \frac{q}{P} \)
• \( \text{time} = \frac{29798.6 \text{ J}}{800 \text{ J/s}} = 37.25 \text{ seconds} \)

Power and energy measurements help you understand the efficiency and effectiveness of devices like microwaves in real-world applications.

Temperature Change Calculation

Temperature change is a common calculation in physics and chemistry when dealing with energy transfers. The temperature change, \( \Delta T \), is calculated as the difference between the final temperature and the initial temperature.
In the given task, we were asked to raise the temperature of water from 24.0°C to 55.0°C. The temperature change can thus be computed as:

• \( \Delta T = 55.0 - 24.0 = 31.0^{\circ} \text{C} \)

Having a precise value for the temperature change is essential because it feeds directly into the heat energy calculation: \( q = m \times c \times \Delta T \).
This calculation forms the basis of understanding how much energy is necessary to reach a desired temperature and is a key part of energy budget planning in practical applications like cooking or heating.