Question

A certain microwave oven delivers 750 . watts \((\mathrm{J} / \mathrm{s})\) of power to a coffee cup containing 50.0 \(\mathrm{g}\) water at \(25.0^{\circ} \mathrm{C}\) . If the wave- length of microwaves in the oven is \(9.75 \mathrm{cm},\) how long does it take, and how many photons must be absorbed, to make the water boil? The specific heat capacity of water is 4.18 \(\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g}\) and assume only the water absorbs the energy of the microwaves

Short answer

It takes approximately 20.9 seconds to make the 50 grams of water in the coffee cup boil and requires about \(7.7 × 10^{28}\) photons to provide the necessary energy.

Step-by-step solution

Find the energy required to heat the water to 100°C

First, we need to find the energy required to heat 50 grams of water from 25°C to 100°C. We can use the formula: \[Q = mc\Delta T\] where Q is the energy required, m is the mass of water, c is the specific heat capacity of water, and \(\Delta T\) is the change in temperature (final temperature - initial temperature). m = 50.0 g, c = 4.18 J/g°C, initial temperature = 25.0°C, final temperature = 100°C \[\Delta T = (100 - 25) °C = 75 °C\] Insert these values into the formula: \[Q = (50.0 \,g)(4.18 \,J/g°C)(75°C)\]

Calculate the energy required to heat the water

Now, calculate the energy Q: \[Q = (50.0)(4.18)(75) = 15675 J\] So, it requires 15,675 joules of energy to make the water boil.

Calculate the time it takes to heat the water

Since we know the microwave oven delivers 750 J/s, we can find the time it takes to deliver the required energy using the formula: \[t = \frac{Q}{P}\] where t is the time, Q is the energy required, and P is the power of the microwave oven. P = 750 J/s, Q = 15675 J Insert these values into the formula: \[t = \frac{15675\,J}{750\,J/s}\]

Find the time it takes

Now, calculate the time t: \[t = \frac{15675}{750} = 20.9\,s\] So, it takes approximately 20.9 seconds to make the water boil.

Calculate the energy of a single photon

Next, let's find the energy of a single photon using the wavelength and the formula: \(E = \frac{hc}{\lambda}\) where E is the photon energy, h is Planck's constant \(6.626 × 10^{-34} Js\), c is the speed of light \(2.998 × 10^8 m/s\), and \(\lambda\) is the wavelength. \[\lambda = 9.75\,cm = 9.75 × 10^{-2}\,m\] Insert these values into the formula: \[E = \frac{(6.626 × 10^{-34} \,Js)(2.998 × 10^8 \,m/s)}{9.75 × 10^{-2} \,m}\]

Find the energy of a single photon

Now, calculate the photon energy E: \[E = 2.0344 × 10^{-24}\,J\] The energy of a single photon is \(2.0344 × 10^{-24}\) joules.

Calculate the number of photons required

Finally, to find how many photons must be absorbed to provide the energy required to heat the water, use the formula: \[N = \frac{Q}{E}\] where N is the number of photons, Q is the energy required, and E is the energy of a single photon. Q = 15675 J, E = \(2.0344 × 10^{-24} J\) Insert these values into the formula: \[N = \frac{15675\,J}{2.0344 × 10^{-24}\,J}\]

Find the number of photons required

Now, calculate the number of photons N: \[N = 7.7 × 10^{28}\] So, it takes approximately \(7.7 × 10^{28}\) photons to make the water boil.

Key concepts

Energy Calculation

Energy calculation is a key concept in understanding how a microwave heats water. This involves determining how much energy is needed to raise the temperature of a specific amount of water. In the context of this exercise, energy calculation allows us to figure out how much microwave power is necessary to bring water to its boiling point.

The formula used is \[Q = mc\Delta T\], where

• \(Q\) is the energy required in joules,
• \(m\) stands for the mass of the substance (in this case, water, measured in grams),
• \(c\) is the specific heat capacity (J/g°C), and
• \(\Delta T\) is the change in temperature (in °C).

This equation tells us that to heat 50 grams of water from 25°C to 100°C, 15,675 joules of energy are needed. To find out how long it takes to heat the water, you divide the energy \(Q\) by the power of the microwave \(P\), given in watts \((J/s)\). Thus, the time taken to make the water boil is calculated with \[t = \frac{Q}{P}\]. This results in 20.9 seconds for our specified microwave.

Using these formulas ensures you precisely understand how energy is used and calculated when it comes to heating.

Specific Heat Capacity

Specific heat capacity is an essential property of substances. It indicates the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. This concept is particularly important when dealing with energy transfer in heating processes.

In this exercise, we dealt with water, which has a specific heat capacity, \(c\), of \(4.18 \, \mathrm{J/g°C}\). It means that 4.18 joules are necessary to increase the temperature of one gram of water by one degree Celsius.

The specific heat capacity helps determine the energy needed to heat a material over a temperature range. For example, heating 50 grams of water from 25°C to 100°C requires 15,675 joules of energy. This value is computed using the specific heat capacity formula \[Q = mc\Delta T\].

Understanding specific heat capacity is crucial because it explains why different substances heat up at different rates. Materials with high specific heat capacity, like water, require more energy to increase in temperature compared to those with lower specific heat capacities.

Photon Energy

Photon energy plays a central role in understanding how microwaves work. Microwaves use electromagnetic radiation to heat substances, and this involves photon energy. A photon is a tiny packet of light or electromagnetic energy. To find the energy of a single photon, we use the formula: \[E = \frac{hc}{\lambda}\]where

• \(E\) is the energy of a photon,
• \(h\) is Planck's constant \((6.626 \times 10^{-34} \, \mathrm{Js})\),
• \(c\) is the speed of light \((2.998 \times 10^8 \, \mathrm{m/s})\), and
• \(\lambda\) is the wavelength \((9.75 \times 10^{-2} \, \mathrm{m})\) in this case.

Calculating these values gives us the energy of a single microwave photon in the oven, yielding \(2.0344 \times 10^{-24} \, \mathrm{J}\). Knowing the energy of a photon allows us to determine how many microwave photons are needed to transfer a specific amount of energy.

Photon energy is thus fundamental for many high-tech applications beyond kitchens, ranging from communication systems to medical techniques.

Wavelength

Wavelength is the distance between two successive peaks of a wave. It is a key characteristic of electromagnetic radiation, including microwaves. In this exercise, the wavelength of the microwave in the oven is given as 9.75 cm.

Wavelength is inversely related to the energy of a photon, meaning the shorter the wavelength, the higher the photon energy. This relationship is expressed via the formula: \[E = \frac{hc}{\lambda}\]. By knowing the wavelength, you can calculate the energy of photons using Planck's constant \(h\) and the speed of light \(c\).

In practical applications, different wavelengths are used for different purposes. A microwave oven, for example, is specifically designed to emit wavelengths that effectively transfer energy to water molecules, making them vibrate and thus, heating the water.

Understanding wavelength helps to tailor the use of electromagnetic radiation in technology—from everyday appliances to scientific research.