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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.

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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.

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Most popular questions from this chapter

For each of the following cases, determine an appropriate characteristic length \(L_{c}\) and the corresponding Biot number \(B i\) that is associated with the transient thermal response of the solid object. State whether the lumped capacitance approximation is valid. If temperature information is not provided, evaluate properties at \(T=300 \mathrm{~K}\). (a) A toroidal shape of diameter \(D=50 \mathrm{~mm}\) and cross-sectional area \(A_{c}=5 \mathrm{~mm}^{2}\) is of thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface of the torus is exposed to a coolant corresponding to a convection coefficient of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) A long, hot AISI 304 stainless steel bar of rectangular cross section has dimensions \(w=3 \mathrm{~mm}\), \(W=5 \mathrm{~mm}\), and \(L=100 \mathrm{~mm}\). The bar is subjected to a coolant that provides a heat transfer coefficient of \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at all exposed surfaces. (c) A long extruded aluminum (Alloy 2024) tube of inner and outer dimensions \(w=20 \mathrm{~mm}\) and \(W=24 \mathrm{~mm}\), respectively, is suddenly submerged in water, resulting in a convection coefficient of \(h=37 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the four exterior tube surfaces. The tube is plugged at both ends, trapping stagnant air inside the tube. (d) An \(L=300-m m\)-long solid stainless steel rod of diameter \(D=13 \mathrm{~mm}\) and mass \(M=0.328 \mathrm{~kg}\) is exposed to a convection coefficient of \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (e) A solid sphere of diameter \(D=12 \mathrm{~mm}\) and thermal conductivity \(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suspended in a large vacuum oven with internal wall temperatures of \(T_{\text {sur }}=20^{\circ} \mathrm{C}\). The initial sphere temperature is \(T_{i}=100^{\circ} \mathrm{C}\), and its emissivity is \(\varepsilon=0.73\). (f) A long cylindrical rod of diameter \(D=20 \mathrm{~mm}\), density \(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\), specific heat \(c_{p}=1750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and thermal conductivity \(k=16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suddenly exposed to convective conditions with \(T_{\infty}=20^{\circ} \mathrm{C}\). The rod is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and reaches a spatially averaged temperature of \(T=100^{\circ} \mathrm{C}\) at \(t=225 \mathrm{~s}\). (g) Repeat part (f) but now consider a rod diameter of \(D=200 \mathrm{~mm}\).

Consider the series solution, Equation \(5.42\), for the plane wall with convection. Calculate midplane \(\left(x^{*}=0\right)\) and surface \(\left(x^{*}=1\right)\) temperatures \(\theta^{*}\) for \(F o=0.1\) and 1 , using \(B i=0.1,1\), and 10 . Consider only the first four eigenvalues. Based on these results, discuss the validity of the approximate solutions, Equations \(5.43\) and \(5.44\).

A wall \(0.12 \mathrm{~m}\) thick having a thermal diffusivity of \(1.5 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) is initially at a uniform temperature of \(85^{\circ} \mathrm{C}\). Suddenly one face is lowered to a temperature of \(20^{\circ} \mathrm{C}\), while the other face is perfectly insulated. (a) Using the explicit finite-difference technique with space and time increments of \(30 \mathrm{~mm}\) and \(300 \mathrm{~s}\), respectively, determine the temperature distribution at \(t=45 \mathrm{~min}\). (b) With \(\Delta x=30 \mathrm{~mm}\) and \(\Delta t=300 \mathrm{~s}\), compute \(T(x, t)\) for \(0 \leq t \leq t_{\mathrm{ss}}\), where \(t_{\mathrm{ss}}\) is the time required for the temperature at each nodal point to reach a value that is within \(1^{\circ} \mathrm{C}\) of the steady-state temperature. Repeat the foregoing calculations for \(\Delta t=75 \mathrm{~s}\). For each value of \(\Delta t\), plot temperature histories for each face and the midplane.

Consider the thick slab of copper in Example 5.12, which is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\) and is suddenly exposed to a net radiant flux of \(3 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\). Use the Finite-Difference Equations/ One-Dimensional/Transient conduction model builder of \(I H T\) to obtain the implicit form of the finite-difference equations for the interior nodes. In your analysis, use a space increment of \(\Delta x=37.5 \mathrm{~mm}\) with a total of 17 nodes (00-16), and a time increment of \(\Delta t=1.2 \mathrm{~s}\). For the surface node 00 , use the finite-difference equation derived in Section 2 of the Example. (a) Calculate the 00 and 04 nodal temperatures at \(t=120 \mathrm{~s}\), that is, \(T(0,120 \mathrm{~s})\) and \(T(0.15 \mathrm{~m}, 120 \mathrm{~s})\), and compare the results with those given in Comment 1 for the exact solution. Will a time increment of \(0.12 \mathrm{~s}\) provide more accurate results? (b) Plot temperature histories for \(x=0,150\), and \(600 \mathrm{~mm}\), and explain key features of your results.

Special coatings are often formed by depositing thin layers of a molten material on a solid substrate. Solidification begins at the substrate surface and proceeds until the thickness \(S\) of the solid layer becomes equal to the thickness \(\delta\) of the deposit. (a) Consider conditions for which molten material at its fusion temperature \(T_{f}\) is deposited on a large substrate that is at an initial uniform temperature \(T_{i}\). With \(S=0\) at \(t=0\), develop an expression for estimating the time \(t_{d}\) required to completely solidify the deposit if it remains at \(T_{f}\) throughout the solidification process. Express your result in terms of the substrate thermal conductivity and thermal diffusivity \(\left(k_{s}, \alpha_{s}\right)\), the density and latent heat of fusion of the deposit \(\left(\rho, h_{s f}\right)\), the deposit thickness \(\delta\), and the relevant temperatures \(\left(T_{f}, T_{i}\right)\). (b) The plasma spray deposition process of Problem \(5.25\) is used to apply a thin \((\delta=2 \mathrm{~mm})\) alumina coating on a thick tungsten substrate. The substrate has a uniform initial temperature of \(T_{i}=300 \mathrm{~K}\), and its thermal conductivity and thermal diffusivity may be approximated as \(k_{s}=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha_{s}=4.0 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. The density and latent heat of fusion of the alumina are \(\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\) and \(h_{s f}=3577 \mathrm{~kJ} / \mathrm{kg}\), respectively, and the alumina solidifies at its fusion temperature \(\left(T_{f}=2318 \mathrm{~K}\right)\). Assuming that the molten layer is instantaneously deposited on the substrate, estimate the time required for the deposit to solidify.
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