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When a swimmer leaves cold water on a warm, breezy day, he experiences a cooling effect. Why?

Short Answer

Expert verified
The cooling effect experienced by a swimmer leaving cold water on a warm, breezy day is primarily due to evaporative cooling. This occurs as heat transfers from the warmer air to the swimmer's colder, wet body, causing the water on the body to evaporate. The breeze enhances this process by removing moisture and increasing the rate of evaporation, and the body responds by increasing blood flow to the skin to facilitate further heat transfer, resulting in a cooler sensation for the swimmer.

Step by step solution

01

Understanding the swimmer's situation

When a swimmer leaves cold water on a warm day, their body is wet and exposed to the air, which is warmer than the water and in motion due to the breeze. This situation creates a temperature difference between the swimmer's body and the surrounding air.
02

Heat transfer from the body to the air

Heat transfer occurs between the swimmer's body (which is colder than the air) and the surrounding warm air. Heat transfers always from hot to cold, so heat from the air transfers to the swimmer's body. This transfer can occur through convection, which is the transfer of heat through a fluid (like air) that is in motion.
03

Evaporative cooling

During the heat transfer process, the water on the swimmer's body starts to evaporate, given that water molecules absorb heat energy from the air and gain enough energy to change from liquid to gas form. This process is called evaporative cooling and is responsible for the cooling effect that the swimmer experiences.
04

Role of the breeze

The breeze plays a significant role in enhancing the evaporative cooling effect. As air flows over the swimmer's wet body, it helps to increase the rate of evaporation by removing the moisture and maintaining a steady supply of warm, dry air. This speeds up the heat transfer process, making the swimmer feel cooler more quickly.
05

Body's response to cooling effect

As the swimmer cools down, their body will also respond by increasing blood flow to the skin. This can stimulate more heat transfer from the body's core to the surface, further enhancing the cooling effect. In conclusion, the cooling effect experienced by a swimmer when they leave cold water on a warm, breezy day is primarily due to evaporative cooling and heat transfer between the swimmer's body and the surrounding warm air, with the breeze enhancing the overall process.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental process that occurs when two objects or substances are at different temperatures. This happens because heat naturally flows from areas of higher temperature to areas with lower temperature. In the scenario of a swimmer leaving cold water, heat transfer begins because of the difference in temperature between the swimmer's body and the warm air.
When the swimmer's body, which is initially cold from the water, comes into contact with the warmer air, heat energy begins to flow into the swimmer. It's important to remember that the swimmer feels cooler not because they're losing heat, but because the mechanisms of heat transfer have complex interactions that change how heat is perceived. The key to understanding this cooling sensation lies within the next few concepts.
Convection
Convection is a type of heat transfer that occurs through the movement of fluids, including air, which acts as a fluid in this context. When the swimmer is standing in the breeze, the moving air plays an important role in heat transfer. Convection transfers heat from the surrounding warm air to the cool surface of the swimmer's wet skin.
This moving fluid facilitates the distribution of heat more efficiently across the skin, especially when there's a breeze. The air molecules that come into contact with the swimmer are warmer, giving their energy to the cooler water on the skin, leading to evaporation. Hence, convection is crucial because it helps drive the heat transfer process by moving warm air constantly over the swimmer, reinforcing the cooling effect experienced.
Temperature Difference
The temperature difference is a key driver of the heat transfer process. It refers to the variance between the temperature of the swimmer's body surface (chilled by the cold water) and the surrounding warmer air.
Initially, the swimmer is cool from the cold water, whereas the air temperature is significantly higher, establishing a temperature difference. This difference causes heat to flow from the air to the swimmer's body, driven by the fundamental principle that heat always shifts from warmer to cooler bodies. The temperature difference is what essentially initiates the entire phenomenon of evaporative cooling and is a necessary condition for the cooling effect to take place.
Role of the Breeze
The breeze has a powerful role in how quickly and effectively evaporative cooling occurs. The presence of a breeze can significantly enhance the rate of evaporation. When air moves over the wet surface of the swimmer's skin, it carries away the moisture droplets that are in the process of turning from liquid to vapor.
The breeze keeps the air around the swimmer constantly fresh and dry. It helps by removing humid air that has been moistened from the evaporation. This constant supply of new, dry air accelerates the evaporation process, allowing more water molecules on the skin to transition into a vapor state. Therefore, the breeze amplifies the overall cooling sensation experienced by the swimmer. The breeze's ability to maintain this cycle is why wind or air movement is so crucial in making the cooling process more effective.

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Most popular questions from this chapter

Assuming that the density of water is \(.9971\left(\mathrm{~g} / \mathrm{cm}^{3}\right)\) at \(25^{\circ} \mathrm{C}\) and that of ice at \(0^{\circ}\) is \(917\left(\mathrm{~g} / \mathrm{cm}^{3}\right)\), what percent of a water jug at \(25^{\circ} \mathrm{C}\) should be left empty so that, if the water freezes, it will just fill the jug?

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A G.T.O. has a 22 gal. cooling system. Suppose you fill it with a \(50-50\) solution by volume of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\), ethylene glycol, and water. At what temperature would freezing become a problem? Assume the specific gravity of ethylene glycol is \(1.115\) and the freezing point depression constant of water is \(\left(1.86 \mathrm{C}^{\circ} / \mathrm{mole}\right)\). You might have placed in methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) instead of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\). If the current cost of ethylene glycol is 12 cents \(/ \mathrm{lb}\) and the cost of methanol is 8 cents \(/ \mathrm{lb}\), how much money would you save by using \(\mathrm{CH}_{3} \mathrm{OH}\) ? And yet, ethylene glycol is the more desirable antifreeze. Why? density \(=(.79 \mathrm{~g} / \mathrm{ml})\) for \(\mathrm{CH}_{3} \mathrm{OH}\) and \(3.785\) liter \(=1\) gallon.

A chemist dissolves \(300 \mathrm{~g}\) of urea in \(1000 \mathrm{~g}\) of water. Urea is \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\). Assuming the solution obeys Raoult's'law, determine the following: a) The vapor pressure of the solvent at \(0^{\circ}\) and \(100^{\circ} \mathrm{C}\) and b) The boiling and freezing point of the solution. The vapor pressure of pure water is \(4.6 \mathrm{~mm}\) and \(760 \mathrm{~mm}\) at \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), respectively and the \(\mathrm{K}_{\mathrm{f}}=(1.86\) \(\mathrm{C} / \mathrm{mole}\) ) and \(\mathrm{K}_{\mathrm{b}}=(52 \mathrm{C} / \mathrm{mole})\)

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