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The figure shows a cross-sectional view of a body covered by a fur layer. A number of convection loops are shown in the fur. The air outside the fur is cold and the body beneath the fur is warm.
Fur is filled with air, breaking it up into many small pockets. Convection is very slow here, because the loops are so small. The low conductivity of air makes fur a very good lightweight insulator.

Some interesting phenomena happen when convection is accompanied by a phase change . It allows us to cool off by sweating, even if the temperature of the surrounding air exceeds body temperature. Heat from the skin is required for sweat to evaporate from the skin, but without air flow, the air becomes saturated and evaporation stops. Air flow caused by convection replaces the saturated air by dry air and evaporation continues.

Calculate the flow of mass during convection: sweat-heat transfer away from the body

The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporate from the body to get rid of all this energy? (This evaporation might occur when a person is sitting in the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer by other methods.)

Strategy

Energy is needed for a phase change ( Q = mL v size 12{Q= ital "mL" rSub { size 8{v} } } {} ). Thus, the energy loss per unit time is

Q t = mL v t = 120  W = 120 J/s.

We divide both sides of the equation by L v size 12{L rSub { size 8{v} } } {} to find that the mass evaporated per unit time is

m t = 120  J/s L v . size 12{ { {m} over {t} } = { {"120"`"J/s"} over {L rSub { size 8{v} } } } } {}

Solution

(1) Insert the value of the latent heat from [link] , L v = 2430  kJ/kg = 2430  J/g size 12{L rSub { size 8{v} } ="2430"`"kJ/kg"="2430"`"J/g"} {} . This yields

m t = 120  J/s 2430  J/g = 0 . 0494  g/s = 2 . 96  g/min. size 12{ { {m} over {t} } = { {"120"`"J/s"} over {"2430"`"J/g"} } =0 "." "044"`"g/s"=2 "." "96"`"g/min"} {}

Discussion

Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz) per hour. If the air is very dry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place in the lungs and breathing passages.

Another important example of the combination of phase change and convection occurs when water evaporates from the oceans. Heat is removed from the ocean when water evaporates. If the water vapor condenses in liquid droplets as clouds form, heat is released in the atmosphere. Thus, there is an overall transfer of heat from the ocean to the atmosphere. This process is the driving power behind thunderheads, those great cumulus clouds that rise as much as 20.0 km into the stratosphere. Water vapor carried in by convection condenses, releasing tremendous amounts of energy. This energy causes the air to expand and rise, where it is colder. More condensation occurs in these colder regions, which in turn drives the cloud even higher. Such a mechanism is called positive feedback, since the process reinforces and accelerates itself. These systems sometimes produce violent storms, with lightning and hail, and constitute the mechanism driving hurricanes.

The figure shows a cumulus cloud in a blue sky.
Cumulus clouds are caused by water vapor that rises because of convection. The rise of clouds is driven by a positive feedback mechanism. (credit: Mike Love)
The figure shows lightning strikes from thunderclouds above an urban area.
Convection accompanied by a phase change releases the energy needed to drive this thunderhead into the stratosphere. (credit: Gerardo García Moretti )
The figure shows some blue-colored icebergs floating in the water beneath snow-capped mountains and a cloudy sky. Some of the icebergs at front are melting.
The phase change that occurs when this iceberg melts involves tremendous heat transfer. (credit: Dominic Alves)

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Can you compute that for me. Ty
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Source:  OpenStax, College physics (engineering physics 2, tuas). OpenStax CNX. May 08, 2014 Download for free at http://legacy.cnx.org/content/col11649/1.2
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