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Chapter 12: Problem 4

How does microwave cooking differ from conventional cooking?

Short Answer

Expert verified
Answer: The primary differences between microwave cooking and conventional cooking include the heating mechanism, cooking time, texture, and safety concerns. Microwave cooking heats food through the use of microwaves that cause water molecules to vibrate and produce heat, while conventional cooking applies direct heat to the food's surface. Microwave cooking is generally faster than conventional methods due to quicker heat penetration. The texture of food cooked in a microwave is typically more moist and lacks the browning or crisping effect of conventional cooking. Safety concerns with microwave cooking include uneven heating and the risk of burns or undercooked portions, necessitating the use of microwave-safe cookware and proper safety precautions.

Step by step solution

01

1. Heating Mechanism

In microwave cooking, food is heated through the use of microwaves, which are a form of electromagnetic radiation. Microwaves penetrate the food and cause water molecules to vibrate, producing heat through the friction between them, which cooks the food. In conventional cooking methods, like using an oven or stovetop, food is cooked through applying direct heat to its surface. This heat then disperses through the food, causing changes in its structure and texture that contribute to the cooking process.
02

2. Cooking Time

Microwave cooking is generally faster than conventional cooking methods. This is because microwaves can penetrate food more quickly, allowing for faster heating of the interior of the food. In contrast, conventional cooking methods involve transferring heat through the outer layers of food, which can take longer and require more energy.
03

3. Texture Differences

The texture of food cooked in a microwave can be different than that of food cooked using conventional methods. Since microwave cooking relies on the vibration of water molecules in the food, it typically results in a more moist texture compared to conventional methods. Additionally, because microwaves cook food from the inside out, they may not provide the same browning or crisping effect that is achieved with conventional cooking methods like baking or frying.
04

4. Safety Concerns

There are some safety concerns unique to microwave cooking. Microwaves can heat food unevenly, creating hot spots that can lead to burns or undercooked portions. This can be particularly problematic when cooking raw meat or other foods that need to reach a certain internal temperature to be considered safe for consumption. It is essential to use microwave-safe cookware and follow all safety precautions when using a microwave, such as stirring food frequently and using a food thermometer to ensure they are cooked to the proper temperature.

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Most popular questions from this chapter

A blackbody plate \(\left(A_{1}=5 \mathrm{~cm}^{2}\right)\) is subjected to a uniform heat flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\) on the bottom, while the top surface is exposed to an ambient surrounding at \(5^{\circ} \mathrm{C}\). The heat transfer coefficient due to natural convection on the plate surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). A radiometer is placed \(1 \mathrm{~m}\) above the plate normal to the direction of viewing from the plate. Determine the irradiation that the radiometer would detect.

What is a blackbody? Does a blackbody actually exist?

A small surface of area \(A=1 \mathrm{~cm}^{2}\) is subjected to incident radiation of constant intensity \(I_{i}=2.2 \times 10^{4} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\) over the entire hemisphere. Determine the rate at which radiation energy is incident on the surface through \((a) 0 \leq \theta\) \(\leq 45^{\circ}\) and (b) \(45 \leq \theta \leq 90^{\circ}\), where \(\theta\) is the angle a radiation beam makes with the normal of the surface.

Daylight and incandescent light may be approximated as a blackbody at the effective surface temperatures of \(5800 \mathrm{~K}\) and \(2800 \mathrm{~K}\), respectively. Determine the wavelength at maximum emission of radiation for each of the lighting sources.

The temperature of the filament of an incandescent lightbulb is \(2500 \mathrm{~K}\). Assuming the filament to be a blackbody, determine the fraction of the radiant energy emitted by the filament that falls in the visible range. Also, determine the wavelength at which the emission of radiation from the filament peaks.
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