/*! 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 Microwave ovens use microwave ra... [FREE SOLUTION] | 91Ó°ÊÓ

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Microwave ovens use microwave radiation to heat food. The energy of the microwaves is absorbed by water molecules in food and then transferred to other components of the food. (a) Suppose that the microwave radiation has a wavelength of \(10 \mathrm{~cm} .\) How many photons are required to heat \(200 \mathrm{~mL}\) of water from 25 to \(75^{\circ} \mathrm{C} ?\) (b) Suppose the microwave's power is \(1000 \mathrm{~W}\) ( 1 watt \(=1\) joule-second \() .\) How long would you have to heat the water in part (a)?

Short Answer

Expert verified
\(a)\) First, calculate the frequency of the radiation: \(\nu = \frac{c}{\lambda} = \frac{3 \times 10^8 m/s}{0.1 m} = 3 \times 10^9 Hz\). Then, find the energy of one photon: \(E = h\nu = 6.63 \times 10^{-34} Js \times 3 \times 10^9 Hz = 1.99 \times 10^{-24} J\). Next, find the energy needed to heat the water: \(Q = mc\Delta T = (0.2 kg)(4.18 \times 10^3 J/kg·°C)(50^{\circ}C) = 4180 J\). Now, calculate the number of photons required: \(\text{Number of photons} = \frac{4180 J}{1.99 \times 10^{-24} J/photon} = 2.1 \times 10^{27} \text{ photons}\). \(b)\) Lastly, find the time taken to heat the water: \(t = \frac{Q}{P} = \frac{4180 J}{1000 W} = 4.18 \ \text{seconds}\).

Step by step solution

01

Calculate the energy of one photon

From the given wavelength (10 cm), first, we need to find the frequency of the microwave radiation using the formula: \(c = \lambda \nu\) Where: c = speed of light \(=3 \times 10^8 m/s\) \(\lambda\) = wavelength \(= 10 cm = 0.1 m\) \(\nu\) = frequency Divide both sides by \(\lambda\): \(\nu = \frac{c}{\lambda}\) Now, we can use Planck's formula to find the energy of one photon: \(E = h\nu\) Where: E = energy of one photon h = Planck's constant \(=6.63 \times 10^{-34} Js\)
02

Calculate the energy needed to heat the water

To find the energy required to heat the given amount of water, we can use the specific heat capacity formula: \(Q = mc\Delta T\) Where: Q = energy required to heat the water m = mass of water c = specific heat capacity of water \(=4.18 \times 10^3 J/kg·°C\) \(\Delta T = T_{final} - T_{initial}\) First, convert 200 mL of water to mass: 1 mL of water = 1 g 200 mL of water = 200 g = 0.2 kg Now, calculate the energy needed to heat the water from 25°C to 75°C: \(\Delta T = 75 - 25 = 50^{\circ}C\)
03

Calculate the number of photons required to heat the water

Now that we have the energy of one photon and the total energy needed to heat the water, we can calculate the number of photons required: \(\text{Number of photons} = \frac{\text{Total energy}}{\text{Energy of one photon}}\)
04

Calculate the time taken to heat the water using the microwave's power

Given the power of the microwave (1000 W = 1000 J/s), we can calculate the time taken to heat the water: \(P = \frac{Q}{t}\) Where: P = power Q = energy t = time Rearrange the formula to solve for time: \(t = \frac{Q}{P}\)

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Key Concepts

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

Photon Energy
Photon energy is a crucial concept when discussing microwave radiation. It refers to the energy carried by a single photon, a particle representing a quantum of light or electromagnetic radiation. Each photon's energy is directly related to its frequency.
  • Higher frequency means higher photon energy.
  • Photon energy is typically measured in joules (J).
To calculate the energy of a photon, we use Planck's formula, which is a vital tool in physics for understanding the energy of electromagnetic waves. The formula is:\[E = hu\]Where:
  • E is the energy of one photon,
  • h is Planck's constant \(6.63 \times 10^{-34} \text{Js}\),
  • u is the frequency of the radiation.
Planck's constant is a very small number denoting the proportionality factor between the energy and frequency of a photon, making it fundamental in quantum mechanics.
Understanding photon energy is essential to determine how effective the microwaves are in transferring energy to heat food.
Specific Heat Capacity
Specific heat capacity is a property of a substance that indicates how much energy is required to raise the temperature of 1 kilogram of that substance by 1 degree Celsius. In our context, we are focused on water's specific heat capacity, which is relatively high at \(4.18 \times 10^3 \text{J/kg} \cdot ^\circ\text{C}\). This high value means water requires a significant amount of energy to increase its temperature.
  • It is why water is an effective medium for heat transfer within food.
  • The energy required to heat water (Q) is found using the formula \(Q = mc\Delta T\).
Here's a breakdown of the formula:
  • Q is the energy required.
  • m is the mass of water, converted from volume (since 1 mL of water is 1 g, 200 mL corresponds to 0.2 kg).
  • c is the specific heat capacity of water.
  • \Delta T is the temperature change, which is the final temperature minus the initial temperature.
Specific heat capacity is a core concept when calculating how microwave energy is utilized to heat foods effectively.
Planck's Formula
Planck's formula is pivotal in calculating the energy of photons. It connects the frequency of electromagnetic radiation to the energy it carries, underpinning much of modern quantum theory.Using Planck's formula:\[E = hu\]you can comprehend how the energy of photons translates to heating capabilities.
  • h (Planck's constant) = 6.63 x \( 10^{-34} \text{Js}\).
  • u is calculated using the formula \( u = \frac{c}{\lambda}\), where \( \lambda \) is the wavelength and c is the speed of light \(3 \times 10^8 \text{m/s}\).
The frequency (\( u \)) of microwaves gives insight into how they efficiently transfer energy to heat food. Understanding how to manipulate and calculate these parameters is essential in determining how many photons are required to produce a specific heating effect.
Microwave Power
Microwave power is another key component when considering the practical application of microwaves for heating purposes. Measured in watts, it indicates the rate at which energy is transferred to the food.
  • A watt (W) is equivalent to one joule per second (J/s), signifying energy flow.
  • In practice, higher power means food heats up faster.
To manage power effectively, the formula \(P = \frac{Q}{t}\) is utilized:
  • P is power (in watts, W).
  • Q is the energy required (in joules, J).
  • t is time (in seconds, s).
This equation helps determine how long it will take to heat food given a specific power level, making it a practical consideration in efficient microwave use. Understanding microwave power is crucial to calculate time efficiency and ensure food is heated adequately in a culinary context.

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Most popular questions from this chapter

Determine which of the following statements are false and correct them. (a) The frequency of radiation increases as the wavelength increases. (b) Electromagnetic radiation travels through a vacuum at a constant speed, regardless of wavelength. (c) Infrared light has higher frequencies than visible light. (d) The glow from a fireplace, the energy within a microwave oven, and a foghorn blast are all forms of electromagnetic radiation.

(a) A green laser pointer emits light with a wavelength of \(532 \mathrm{nm}\). What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of \(532-\mathrm{nm}\) photons. What is the energy gap between the ground state and excited state in the laser material?

Bohr's model can be used for hydrogen-like ions-ions that have only one electron, such as \(\mathrm{He}^{+}\) and \(\mathrm{Li}^{2+} .\) (a) Why is the Bohr model applicable to \(\mathrm{Li}^{2+}\) ions but not to neutral Li atoms? (b) The ground-state energies of \(\mathrm{B}^{4+}, \mathrm{C}^{5+},\) and \(\mathrm{N}^{6+}\) are tabulated as follows: By examining these numbers, propose a relationship between the ground-state energy of hydrogen-like systems and the nuclear charge, \(Z\). (Hint: Divide by the ground-state energy of hydrogen \(\left.-2.18 \times 10^{-18} \mathrm{~J}\right)\) (c) Use the relationship you derive in part (b) to predict the ground-state energy of the \(\mathrm{Be}^{3+}\) ion.

State where in the periodic table these elements appear: (a) elements with the valence-shell electron configuration \(n s^{2} n p^{5}\) (b) elements that have three unpaired \(p\) electrons (c) an element whose valence electrons are \(4 s^{2} 4 p^{1}\) (d) the \(d\) -block elements [Section 6.9\(]\)

Using the periodic table as a guide, write the condensed electron configuration and determine the number of unpaired electrons for the ground state of (a) \(\mathrm{Cl},(\mathbf{b}) \mathrm{Al},(\mathbf{c}) \mathrm{Zr},(\mathbf{d})\) As, (e) \(\mathrm{Sb},(\mathbf{f}) \mathrm{W}\)

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