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Chapter 7: Problem 148

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

Expert verified
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

01

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: \[Q = (50.0 \,g)(4.18 \,J/g°C)(75°C)\]
02

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.
03

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: \[t = \frac{15675\,J}{750\,J/s}\]
04

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.
05

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: \[E = \frac{(6.626 × 10^{-34} \,Js)(2.998 × 10^8 \,m/s)}{9.75 × 10^{-2} \,m}\]
06

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.
07

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: \[N = \frac{15675\,J}{2.0344 × 10^{-24}\,J}\]
08

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.

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