/*! 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 3 A microwave oven operates on the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 3

A microwave oven operates on the principle that application of a high- frequency field causes electrically polarized molecules in food to oscillate. The net effect is a nearly uniform generation of thermal energy within the food. Consider the process of cooking a slab of beef of thickness \(2 L\) in a microwave oven and compare it with cooking in a conventional oven, where each side of the slab is heated by radiation. In each case the meat is to be heated from \(0^{\circ} \mathrm{C}\) to a minimum temperature of \(90^{\circ} \mathrm{C}\). Base your comparison on a sketch of the temperature distribution at selected times for each of the cooking processes. In particular, consider the time \(t_{0}\) at which heating is initiated, a time \(t_{1}\) during the heating process, the time \(t_{2}\) corresponding to the conclusion of heating, and a time \(t_{3}\) well into the subsequent cooling process.

Short Answer

Expert verified
In conclusion, the main difference between the two cooking processes lies in the uniformity of temperature distribution during the cooking process. A microwave oven provides nearly uniform heating throughout the slab of beef, leading to more even cooking, whereas a conventional oven has a non-uniform temperature distribution due to surface heating and heat conduction. As a result, the slab of beef cooked in a microwave oven has a more consistent temperature throughout, while the one cooked in a conventional oven has a hotter surface and potentially cooler interior.

Step by step solution

01

Understanding cooking processes in microwave and conventional ovens

A microwave oven utilizes high-frequency electromagnetic waves to cause polarized water molecules in the food to oscillate, which produces thermal energy to cook the food. The heating is nearly uniform throughout the food. In contrast, a conventional oven relies on the radiation from its heating elements to heat the food's surfaces. The heat then conducts through the food, moving from the hotter surfaces to the cooler interior. This heating process is less uniform compared to the microwave oven.
02

Temperature distribution at \(t_0\)

At time \(t_0\), both cooking processes have just begun, and the initial temperature of the slab of beef is \(0^{\circ}\mathrm{C}\). Hence, the temperature distribution is uniform at \(0^{\circ}\mathrm{C}\) for both microwave and conventional ovens.
03

Temperature distribution at \(t_1\)

At time \(t_1\), during the cooking process: - In a microwave oven, the temperature distribution remains nearly uniform throughout the cooking process due to the direct interaction of microwaves with polarized water molecules in the slab of beef. - In a conventional oven, the slab of beef's surface is exposed to the radiation and gets heated. The heat then spreads from the hotter surface to the cooler interior. Therefore, the temperature distribution is not uniform as the slab's interior remains cooler than its surface.
04

Temperature distribution at \(t_2\)

At time \(t_2\), when the cooking process is complete: - In a microwave oven, since the heating is nearly uniform throughout the slab of beef, the temperature would reach around \(90^{\circ}\mathrm{C}\) uniformly at the end of the cooking process. - In a conventional oven, however, the temperature distribution is not uniform. The surface would have reached or exceeded \(90^{\circ}\mathrm{C}\), but the interior might still be cooler than \(90^{\circ}\mathrm{C}\).
05

Temperature distribution at \(t_3\)

At time \(t_3\), well into the cooling process: - Both microwave and conventional oven-cooked slabs of beef go through a cooling process, and the temperature drops in both cases. However, the exact temperature distribution during the cooling phase depends on a variety of factors, such as the cooling environment, whether the food is covered, etc. In conclusion, the main difference between the two cooking processes lies in the uniformity of temperature distribution during the cooking process, which is significantly better in a microwave oven compared to a conventional oven. This results in overall more even cooking of the slab of beef in a microwave oven than in a conventional oven.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Microwave Cooking
Microwave cooking uses electromagnetic waves, typically at a high frequency, to heat food. These waves cause water molecules and other polarized substances within the food to oscillate rapidly. As a result, these molecules bump into each other, generating heat uniformly across the entire dish.
This method is popular due to its efficiency and speed. Unlike traditional cooking methods, microwaves penetrate and excite molecules throughout the bulk of the food, rather than just heating the surface. This is why food can often cook faster with a microwave than with other methods. The energy conversion in microwaves is nearly instant, so the food starts cooking right away when the oven is turned on.
  • Heats quickly and uniformly
  • Uses electromagnetic waves
  • Excites water molecules
  • Great for even heating
Microwave cooking is excellent for speeding up mealtime preparations. Its main advantage is that it heats food from the inside out, unlike ovens which often take longer to reach the desired internal temperature of the food.
Convection Cooking
Convection cooking occurs in appliances like conventional ovens and relies on the circulation of hot air or other fluids. In a convection oven, fans circulate the hot air around the food. This helps to distribute heat more evenly compared to regular ovens that rely solely on gravity.
However, convection cooking primarily heats the surfaces that are exposed. The heat is then conducted inward through the food. This leads to different temperature layers, especially if the food has a significant thickness. The outer layers reach the desired temperature first, while the inside might take longer to cook thoroughly.
  • Relies on hot air circulation
  • Heats surfaces first
  • Needs time for internal heat conduction
  • Fans help distribute heat evenly
Convection ovens excel in cooking dishes evenly compared to standard ovens but still require time for the heat to penetrate the food's core.
Temperature Distribution
Temperature distribution refers to how heat is spread throughout the item being cooked. In microwave cooking, the temperature distribution is almost uniform due to the direct interaction of microwaves with the food's molecules. This means that the heat is evenly spread throughout the food.
In contrast, during convection cooking, the temperature at the surface can be much higher than the temperature at the center because the heat transfers through conduction after the surface absorbs it. This can result in uneven cooking, with the exterior potentially overcooked while the interior is undercooked if not managed properly.
  • Microwaves offer nearly uniform temperature distribution
  • Convection leads to temperature gradients
  • Surface heats first in conventional methods
  • Important factor for desired cooking results
Understanding temperature distribution is essential for achieving the best results in cooking, as it determines the overall doneness and texture of the food.
Thermal Energy Generation
Thermal energy generation is at the heart of the cooking process. In a microwave oven, thermal energy is generated by causing water molecules to oscillate and produce heat instantaneously within the food. This approach makes the microwave highly efficient as it directly converts electromagnetic energy to thermal energy right where it's needed.
On the other hand, in conventional ovens, thermal energy is generated by heating elements. It is then transferred to the food through a combination of radiation, convection, and conduction. This process is naturally slower because the energy must first transfer from the heating elements to the air, then to the food's surface, and finally conduct to the food's core.
  • Microwaves generate thermal energy quickly
  • Conventional ovens use a slower heat transfer process
  • Energy generation affects cooking time and efficiency
  • Central to understanding different cooking methods
Recognizing how thermal energy is generated in different cooking appliances helps in selecting the right tool for the cooking task, ensuring efficiency and time management.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a material processing experiment conducted aboard the space shuttle, a coated niobium sphere of \(10-\mathrm{mm}\) diameter is removed from a furnace at \(900^{\circ} \mathrm{C}\) and cooled to a temperature of \(300^{\circ} \mathrm{C}\). Although properties of the niobium vary over this temperature range, constant values may be assumed to a reasonable approximation, with \(\rho=8600 \mathrm{~kg} / \mathrm{m}^{3}, c=290 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=\) \(63 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) If cooling is implemented in a large evacuated chamber whose walls are at \(25^{\circ} \mathrm{C}\), determine the time required to reach the final temperature if the coating is polished and has an emissivity of \(\varepsilon=0.1\). How long would it take if the coating is oxidized and \(\varepsilon=0.6\) ? (b) To reduce the time required for cooling, consideration is given to immersion of the sphere in an inert gas stream for which \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=\) \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Neglecting radiation, what is the time required for cooling? (c) Considering the effect of both radiation and convection, what is the time required for cooling if \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\varepsilon=0.6\) ? Explore the effect on the cooling time of independently varying \(h\) and \(\varepsilon\).

Plasma spray-coating processes are often used to provide surface protection for materials exposed to hostile environments, which induce degradation through factors such as wear, corrosion, or outright thermal failure. Ceramic coatings are commonly used for this purpose. By injecting ceramic powder through the nozzle (anode) of a plasma torch, the particles are entrained by the plasma jet, within which they are then accelerated and heated. During their time-in-fbht, the ceramic particles must be heated to their melting point and experience complete conversion to the liquid state. The coating is formed as the molten droplets impinge (splat) on the substrate material and experience rapid solidification. Consider conditions for which spherical alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right.\) ) particles of diameter \(D_{p}=50 \mu \mathrm{m}\), density \(\rho_{p}=\) \(3970 \mathrm{~kg} / \mathrm{m}^{3}\), thermal conductivity \(k_{p}=10.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and specific heat \(c_{p}=1560 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) are injected into an arc plasma, which is at \(T_{\infty}=10,000 \mathrm{~K}\) and provides a coefficient of \(h=30,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for convective heating of the particles. The melting point and latent heat of fusion of alumina are \(T_{\text {mp }}=2318 \mathrm{~K}\) and \(h_{s f}=3577 \mathrm{~kJ} / \mathrm{kg}\), respectively. (a) Neglecting radiation, obtain an expression for the time-in-flight, \(t_{i-f}\), required to heat a particle from its initial temperature \(T_{i}\) to its melting point \(T_{\text {mp }}\), and, once at the melting point, for the particle to experience complete melting. Evaluate \(t_{i-f}\) for \(T_{i}=300 \mathrm{~K}\) and the prescribed heating conditions. (b) Assuming alumina to have an emissivity of \(\varepsilon_{p}=0.4\) and the particles to exchange radiation with large surroundings at \(T_{\text {sur }}=300 \mathrm{~K}\), assess the validity of neglecting radiation.

Standards for firewalls may be based on their thermal response to a prescribed radiant heat flux. Consider a \(0.25\)-m-thick concrete wall \(\left(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is at an initial temperature of \(T_{i}=25^{\circ} \mathrm{C}\) and irradiated at one surface by lamps that provide a uniform heat flux of \(q_{s}^{\prime \prime}=10^{4} \mathrm{~W} / \mathrm{m}^{2}\). The absorptivity of the surface to the irradiation is \(\alpha_{s}=1.0\). If building code requirements dictate that the temperatures of the irradiated and back surfaces must not exceed \(325^{\circ} \mathrm{C}\) and \(25^{\circ} \mathrm{C}\), respectively, after \(30 \mathrm{~min}\) of heating, will the requirements be met?

Stainless steel (AISI 304) ball bearings, which have uniformly been heated to \(850^{\circ} \mathrm{C}\), are hardened by quenching them in an oil bath that is maintained at \(40^{\circ} \mathrm{C}\). The ball diameter is \(20 \mathrm{~mm}\), and the convection coefficient associated with the oil bath is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If quenching is to occur until the surface temperature of the balls reaches \(100^{\circ} \mathrm{C}\), how long must the balls be kept in the oil? What is the center temperature at the conclusion of the cooling period? (b) If 10,000 balls are to be quenched per hour, what is the rate at which energy must be removed by the oil bath cooling system in order to maintain its temperature at \(40^{\circ} \mathrm{C}\) ?

In a manufacturing process, long rods of different diameters are at a uniform temperature of \(400^{\circ} \mathrm{C}\) in a curing oven, from which they are removed and cooled by forced convection in air at \(25^{\circ} \mathrm{C}\). One of the line operators has observed that it takes \(280 \mathrm{~s}\) for a \(40-\mathrm{mm}\) diameter rod to cool to a safe-to-handle temperature of \(60^{\circ} \mathrm{C}\). For an equivalent convection coefficient, how long will it take for an 80 -mm-diameter rod to cool to the same temperature? The thermophysical properties of the rod are \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Comment on your result. Did you anticipate this outcome?
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Physics of Motion

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electricity

Read Explanation

Fluids

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.