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Why is a dry, fluffy coat a better insulator than the same coat when it is wet?

Short Answer

Expert verified
Answer: A dry, fluffy coat is a better insulator because it contains numerous air pockets that resist the flow of heat. When the coat becomes wet, water molecules fill the air pockets and reduce the coat's insulating properties, as water is a better conductor of heat than air. The presence or absence of trapped air pockets significantly affects the insulating properties of the coat.

Step by step solution

01

Understanding insulation

Insulation is the process by which a material resists the flow of heat. Good insulators help keep heat in during cold temperatures and keep heat out during warm temperatures. The property of insulation is often determined by the composition of a material and how its structure affects heat transfer.
02

Explaining the role of air pockets and fluffy materials

Fluffy materials with many air pockets, such as the dry, fluffy coat, are excellent insulators because air itself is a poor conductor of heat. The trapped air pockets within the fluffy coat create a barrier that resists heat transfer, causing the coat to act as an effective insulator.
03

Discussing the impact of water on the coat's insulation properties

When the same fluffy coat becomes wet, the water molecules can fill the air pockets, reducing the space between fibers of the coat. Water is a better conductor of heat than air, which means it transfers heat more efficiently than trapped air. Consequently, when the air pockets are filled with water, the coat loses its insulating properties due to the increased heat transfer.
04

Comparing dry and wet coats

In summary, a dry, fluffy coat provides better insulation because it contains numerous air pockets that resist the flow of heat. On the other hand, when the coat becomes wet, the water molecules replace the insulating air pockets, making the coat less effective as an insulator. The key difference between the two situations lies in the presence or absence of trapped air pockets, which significantly affects the insulating properties of the coat.

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Most popular questions from this chapter

In one of your rigorous workout sessions, you lost \(150 \mathrm{~g}\) of water through evaporation. Assume that the amount of work done by your body was \(1.80 \cdot 10^{5} \mathrm{~J}\) and that the heat required to evaporate the water came from your body. a) Find the loss in internal energy of your body, assuming the latent heat of vaporization is \(2.42 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}\). b) Determine the minimum number of food calories that must be consumed to replace the internal energy lost (1 food calorie \(=4186\) J).

Suppose you mix 7.00 L of water at \(2.00 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) with \(3.00 \mathrm{~L}\) of water at \(32.0^{\circ} \mathrm{C}\); the water is insulated so that no energy can flow into it or out of it. (You can achieve this, approximately, by mixing the two fluids in a foam cooler of the kind used to keep drinks cool for picnics.) The \(10.0 \mathrm{~L}\) of water will come to some final temperature. What is this final temperature?

The human body transports heat from the interior tissues, at temperature \(37.0^{\circ} \mathrm{C},\) to the skin surface, at temperature \(27.0^{\circ} \mathrm{C},\) at a rate of \(100 . \mathrm{W}\). If the skin area is \(1.5 \mathrm{~m}^{2}\) and its thickness is \(3.0 \mathrm{~mm}\), what is the effective thermal conductivity, \(\kappa,\) of skin?

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - \(\mathrm{L}\) plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of \(6.00 \mathrm{~cm}\), a height of \(12.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm}\). Use \(205 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of aluminum. Assume that the 2.00 - \(\mathrm{L}\) bottle of soda has a diameter of \(10.0 \mathrm{~cm}\), a height of \(25.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm} .\) Use \(0.100 \mathrm{~W} /(\mathrm{mK})\) as the thermal conductivity of plastic.

Arthur Clarke wrote an interesting short story called "A Slight Case of Sunstroke." Disgruntled football fans came to the stadium one day equipped with mirrors and were ready to barbecue the referee if he favored one team over the other. Imagine the referee to be a cylinder filled with water of mass \(60.0 \mathrm{~kg}\) at \(35.0^{\circ} \mathrm{C}\). Also imagine that this cylinder absorbs all the light reflected on it from 50,000 mirrors. If the heat capacity of water is \(4.20 \cdot 10^{3} \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right),\) how long will it take to raise the temperature of the water to \(100 .{ }^{\circ} \mathrm{C}\) ? Assume that the Sun gives out \(1.00 \cdot 10^{3} \mathrm{~W} / \mathrm{m}^{2},\) the dimensions of each mirror are \(25.0 \mathrm{~cm}\) by \(25.0 \mathrm{~cm},\) and the mirrors are held at an angle of \(45.0^{\circ}\)

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