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Chapter 10: Problem 111

When a person has a severe fever, one therapy used to reduce the fever is an "alcohol rub." Explain how the evaporation of alcohol from a person's skin removes heat energy from the body.

Short Answer

Expert verified
The evaporation of alcohol from a person's skin removes heat energy from the body due to three key factors: the absorption of latent heat of vaporization, heat conduction between the skin and alcohol, and the faster evaporation rate of alcohol compared to water. As the alcohol evaporates, it absorbs heat energy from the body, causing the body temperature to decrease and, therefore, reducing the fever.

Step by step solution

01

Understand evaporation and its cooling effect

Evaporation is the process where a liquid turns into its vapor (gaseous) form at a temperature below its boiling point. When a liquid evaporates, it absorbs the latent heat of vaporization from its surroundings, which causes the removal of heat energy and results in a cooling effect. In the case of an alcohol rub, the alcohol evaporates at room/body temperature, taking away the heat energy from the person's skin.
02

Heat transfer from the body to alcohol

Heat energy is transferred from the body to the alcohol by conduction and convection. When an alcohol rub is applied to a person's skin, the alcohol absorbs the heat energy from the body by conduction. This process of heat transfer occurs at the molecular level as the warmer molecules of the body lose energy to the cooler molecules of the alcohol.
03

Alcohol's faster evaporation rate

Due to its lower boiling point, alcohol evaporates faster than water. This quick-evaporation property is essential in achieving the desired cooling effect. When alcohol is applied to the skin, it rapidly evaporates and absorbs the heat energy from the body's surface, resulting in a decrease in body temperature.
04

How alcohol rub reduces fever

Now that we understand the physics behind evaporation and its cooling effect, we can explain how an alcohol rub reduces fever. When a person is experiencing a fever, their body temperature is higher than normal. An alcohol rub is applied to the skin, causing the evaporation of alcohol. This evaporation process removes heat energy from the body, causing the body temperature to decrease. In conclusion, the evaporation of alcohol from a person's skin removes heat energy due to the absorption of latent heat of vaporization, heat conduction between the skin and alcohol, and the faster evaporation rate of alcohol. The person's body temperature is lowered, helping to reduce the fever.

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Most popular questions from this chapter

Explain why water forms into beads on a waxed car finish.

The unit cell of MgO is shown below. Does \(\mathrm{MgO}\) have a structure like that of \(\mathrm{NaCl}\) or \(\mathrm{ZnS} ?\) If the density of \(\mathrm{MgO}\) is \(3.58 \mathrm{~g} / \mathrm{cm}^{3}\), estimate the radius (in centimeters) of the \(\mathrm{O}^{2-}\) anions and the \(\mathrm{Mg}^{2+}\) cations.

Iodine, like most substances, exhibits only three phases: solid, liquid, and vapor. The triple point of iodine is at 90 torr and \(115^{\circ} \mathrm{C}\). Which of the following statements concerning liquid \(\mathrm{I}_{2}\) must be true? Explain your answer. a. \(\mathrm{I}_{2}(l)\) is more dense than \(\mathrm{I}_{2}(g)\). b. \(\mathrm{I}_{2}(l)\) cannot exist above \(115^{\circ} \mathrm{C}\). c. \(\mathrm{I}_{2}(l)\) cannot exist at 1 atmosphere pressure. d. \(\mathrm{I}_{2}(l)\) cannot have a vapor pressure greater than 90 torr. e. \(\mathrm{I}_{2}(l)\) cannot exist at a pressure of 10 torr.

The \(\mathrm{CsCl}\) structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 67 ). Given that the density of cesium chloride is \(3.97 \mathrm{~g} / \mathrm{cm}^{3}\), and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent \(\mathrm{Cs}^{+}\) and \(\mathrm{Cl}^{-}\) ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of \(\mathrm{Cs}^{+}\) is \(169 \mathrm{pm}\), and the ionic radius of \(\mathrm{Cl}^{-}\) is \(181 \mathrm{pm}\).

A \(20.0-\mathrm{g}\) sample of ice at \(-10.0^{\circ} \mathrm{C}\) is mixed with \(100.0 \mathrm{~g}\) water at \(80.0^{\circ} \mathrm{C}\). Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are \(2.03\) and \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the enthalpy of fusion for ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\).
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