/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 105 A microwave oven operating at \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A microwave oven operating at \(1.22 \times 10^{8} \mathrm{nm}\) is used to heat \(150 \mathrm{~mL}\) of water (roughly the volume of a tea cup) from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). Calculate the number of photons needed if 92.0 percent of microwave energy is converted to the thermal energy of water.

Short Answer

Expert verified
Approximately \(3.34 \times 10^{28}\) photons are needed.

Step by step solution

01

Calculate the Energy Needed to Heat Water

To find the energy required to heat the water, use the formula:\[ q = mc\Delta T \]where:- \( m = 150 \text{ g} \) (since 1 mL of water has a mass of 1 g)- \( c = 4.18 \text{ J/g°C} \) (specific heat capacity of water)- \( \Delta T = 100 - 20 = 80 \text{°C} \)Substitute the values into the formula:\[ q = 150 \times 4.18 \times 80 = 50160 \text{ J} \]
02

Adjust for Microwave Efficiency

Since only 92.0% of the microwave's energy is converted into thermal energy, we must adjust the energy calculation:\[ \text{Energy from microwave} = \frac{50160}{0.92} = 54521.74 \text{ J} \]
03

Calculate Energy of a Single Photon

The energy of a single photon is given by the formula:\[ E = \frac{hc}{\lambda} \]where:- \( h = 6.626 \times 10^{-34} \text{ Js} \) (Planck's constant)- \( c = 3 \times 10^8 \text{ m/s} \) (speed of light)- \( \lambda = 1.22 \times 10^8 \text{ nm} \) Convert \( \lambda \) to meters:\[ \lambda = 1.22 \times 10^{-1} \text{ m} \]Substitute into the formula:\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.22 \times 10^{-1}} = 1.63 \times 10^{-24} \text{ J/photon} \]
04

Calculate the Number of Photons Needed

To find the total number of photons required, divide the total energy by the energy per photon:\[ N = \frac{54521.74}{1.63 \times 10^{-24}} \approx 3.34 \times 10^{28} \text{ photons} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Conversion in Microwaves
Microwave ovens are common household appliances known for their ability to quickly heat food and beverages. This happens through a process called energy conversion. In simple terms, energy conversion involves transforming one form of energy into another. In the case of microwaves, electrical energy is converted into microwave energy that then interacts with water molecules in food to generate heat.

This process is crucial, as it dictates the heating efficiency of the microwave oven. Most of the microwave's energy successfully converts into thermal energy, but some is lost in the form of heat dissipation or reflection. For example, in this exercise, only 92% of the microwave's energy effectively heats the water. Understanding this concept of energy conversion helps in evaluating the effectiveness and energy consumption of a microwave oven in practical applications.
Thermal Energy and Heating Water
Thermal energy refers to the internal energy present in a system due to its temperature. When you heat water, you are essentially increasing its thermal energy. The formula used to calculate the energy needed to change the temperature of a substance is given by:
  • \( q = mc\Delta T \)
In this equation, \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

For water, the specific heat capacity is 4.18 J/g°C, which means it can absorb a significant amount of heat before its temperature rises. When you apply this formula to heat 150 mL (or grams) of water from 20°C to 100°C, you calculate that 50,160 Joules of thermal energy is necessary. This is the energy required to increase the water’s thermal store, using the energy conversion pathway previously mentioned.
Understanding Planck's Constant
Planck's constant is a fundamental physical constant that plays a vital role in quantum mechanics. It is the proportionality factor between the minimal increment of energy, a photon, and the frequency of its associated electromagnetic wave. Represented as \( h \), Planck's constant has the value \( 6.626 \times 10^{-34} \text{ Js} \).

This constant is critical when calculating the energy of a photon. Using the equation \( E = \frac{hc}{\lambda} \), where \( E \) is the energy of the photon, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength, you can determine just how much energy a single photon carries. This constant suggests that energy and frequency are directly related, emphasizing the discrete nature of electromagnetic radiation, including microwaves seen in everyday appliances.
Photon Calculation in Microwave Heating
Calculating the number of photons involved in heating processes, such as in a microwave, involves quantifying how many individual packets of energy (photons) are needed. A single photon's energy is determined using the equation \( E = \frac{hc}{\lambda} \), which you've learned previously.

In our exercise, with a wavelength of \( 1.22 \times 10^8 \text{ nm} \), the energy of one photon comes out to \( 1.63 \times 10^{-24} \text{ J} \). To determine how many photons are needed to deliver the sufficient energy (54,521.74 Joules) to heat the water, you can use:
  • \( N = \frac{\text{Total Energy}}{\text{Energy per Photon}} \)
This leads to required photon count calculated as approximately \( 3.34 \times 10^{28} \) photons. Understanding photon calculations helps bridge the gap between microscopic quantum events and macroscopic results, such as a hot cup of tea from microwave heating.

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Most popular questions from this chapter

The work function of potassium is \(3.68 \times 10^{-19} \mathrm{~J}\). (a) What is the minimum frequency of light needed to eject elections from the metal? (b) Calculate the kinetic energy of the ejected electrons when light of frequency equal to \(8.62 \times 10^{14} \mathrm{~s}^{-1}\) is used for irradiation.

What is the maximum number of electrons in an atom that can have the following quantum numbers? Specify the orbitals in which the electrons would be found. (a) \(n=2, m_{s}=+\frac{1}{2} ;\) (b) \(n=4, m_{\ell}=+1 ;\) (c) \(n=3\), \(\ell=2 ;\) (d) \(n=2, \ell=0, m_{s}=-\frac{1}{2} ;\) (e) \(n=4, \ell=3,\) \(m_{\ell}=-2\)

Indicate the total number of (a) \(p\) electrons in \(\mathrm{N}(Z=7)\) (b) \(s\) electrons in \(\operatorname{Si}(Z=14) ;\) and \((\mathrm{c}) 3 d\) electrons in \(\mathrm{S}(Z=16)\)

Which orbital in each of the following pairs is lower in energy in a many- electron atom? (a) \(2 s, 2 p ;\) (b) \(3 p\), \(3 d ;\) (c) \(3 s, 4 s ;\) (d) \(4 d, 5 f\)

When an electron makes a transition between energy levels of a hydrogen atom, there are no restrictions on the initial and final values of the principal quantum number \(n\). However, there is a quantum mechanical rule that restricts the initial and final values of the orbital angular momentum \(\ell .\) This is the selection rule, which states that \(\Delta \ell=\pm 1,\) that is, in a transition, the value of \(\ell\) can only increase or decrease by one. According to this rule, which of the following transitions are allowed: (a) \(1 s \longrightarrow 2 s\), (b) \(2 p \longrightarrow 1 s,\) (c) \(1 s \longrightarrow 3 d\), (d) \(3 d \longrightarrow 4 f\), (e) \(4 d \longrightarrow 3 s ?\) For hydrogenlike ions, that is, ions containing only one electron, Equation (7.5) is modified as follows: \(E_{n}=-R_{\mathrm{H}} Z^{2}\left(1 / n^{2}\right),\) where \(Z\) is the atomic number of

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