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Chapter 19: Problem 13

When you blow hard on your hand, it feels cool, but when you breathe softly, it feels warm. Why?

Short Answer

Expert verified
Answer: The difference in perception of temperature when blowing hard and breathing softly on your hand is due to the different rates of heat transfer caused by the difference in air speed. Faster air speed when blowing hard leads to a faster evaporation rate and higher heat transfer rate, making your hand feel cooler. Conversely, slower air speed when breathing softly results in a lower evaporation rate and slower heat transfer rate, making your hand feel warmer.

Step by step solution

01

Describe the perception of temperature

When you blow hard on your hand, it feels cool, and when you breathe softly, it feels warm. This difference is related to the rate of evaporation and heat transfer between the air and your hand.
02

Discuss the role of evaporation

Evaporation is a process in which liquid turns into vapor. In our bodies, sweat is produced by sweat glands, and when it evaporates, it absorbs heat from the skin, providing a cooling effect. The faster the sweat evaporates, the cooler your hand will feel.
03

Compare heat transfer rates with different air speeds

When you blow on your hand with a high speed, the air moves over your skin more quickly, leading to a higher evaporation rate. This faster evaporation rate increases the heat transfer rate from your hand to the air, making your hand feel cooler. When you breathe softly on your hand, the air speed is slower, resulting in a lower evaporation rate. Therefore, the heat transfer rate from your hand to the air is also slower, allowing the warm moisture in your breath to accumulate near your skin and make your hand feel warmer.
04

Explain the difference in perception of temperature

The difference in perception of temperature when blowing hard and breathing softly on your hand is due to the different rates of heat transfer caused by the difference in air speed. When blowing hard on your hand, the faster air speed leads to a faster evaporation rate, which causes a higher heat transfer rate and makes your hand feel cooler. When breathing softly, the slower air speed and lower evaporation rate result in a slower heat transfer rate, making your hand feel warmer.

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Most popular questions from this chapter

A quantity of liquid water comes into equilibrium with the air in a closed container, without completely evaporating, at a temperature of \(25.0^{\circ} \mathrm{C} .\) How many grams of water vapor does a liter of the air contain in this situation? The vapor pressure of water at \(25.0^{\circ} \mathrm{C}\) is \(3.1690 \mathrm{kPa}\).

Liquid bromine \(\left(\mathrm{Br}_{2}\right)\) is spilled in a laboratory accident and evaporates. Assuming the vapor behaves as an ideal gas, with a temperature \(20.0^{\circ} \mathrm{C}\) and a pressure of \(101.0 \mathrm{kPa}\), find its density.

a) What is the root-mean-square speed for a collection of helium- 4 atoms at \(300 . \mathrm{K} ?\) b) What is the root-mean-square speed for a collection of helium- 3 atoms at 300 . K?

One mole of an ideal gas, at a temperature of \(0^{\circ} \mathrm{C}\), is confined to a volume of \(1.0 \mathrm{~L}\). The pressure of this gas is a) \(1.0 \mathrm{~atm}\). b) 22.4 atm. c) \(1 / 22.4 \mathrm{~atm}\) d) \(11.2 \mathrm{~atm}\).

6.00 liters of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergo the following processes: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the pressure, volume, and temperature of the gas in states 2 and \(3 .\) How many moles of the gas are there?
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