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Describes process of and factors influencing liquid to gas state change

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Sauna Survival

Sauna Survival

Credit: Olaf Tausch
Source: http://commons.wikimedia.org/wiki/File:NaturTherme_Templin_Sauna_07.jpg
License: CC BY-NC 3.0

A summer day with a temperature of 40oC (104oF) would be unpleasantly warm. A high of 50oC (122oF) would be positively sweltering. Inside a typical sauna, as shown in the image above, you might find temperatures of 70-80oC (158-176oF), or sometimes even higher. The human body requires a steady temperature of about 37oC (98.6oF) in order to function properly. How can humans survive under such hot conditions? Our ability to sweat provides a major part of the answer.

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  • The act of sweating alone is not enough to cool the body. However, changing water from a liquid to a gas is an endothermic process. As the sweat evaporates, energy is absorbed from the surroundings (your skin), resulting in a decrease in temperature.
  • Credit: jeffrey montes
    Source: http://www.flickr.com/photos/jm-photography/4728646934/
    License: CC BY-NC 3.0

    Sweating allows the air to cool the skin by evaporating the water [Figure2]


  • On humid days, the air already contains a lot of water vapor. As a result, evaporation is not as rapid, and sweating is a less effective way to keep cool. This is why humid days often seem hotter than a dryer day at the same temperature.
  • Saunas are quite common, so clearly the human body can survive temperatures of 70-80oC, at least for a short while. What about 100oC (the boiling point of water)? What about even higher temperatures? Experiments done by Charles Blagden in the late 1700s provide some answers. Learn more about them in the following video: 


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With the links below, learn more about the energetic changes associated with the evaporation of water. Then answer the following questions.

  1. When exercising, an adult human will normally produce about one liter of sweat per hour. The heat of vaporization of water at body temperature is about 43.5 kJ/mol. (Note that this value is slightly different from DHvap at the boiling point of water.) Using this value, calculate the amount of energy required to vaporize 1.00 liter of water at body temperature, assuming a density of 1.00 g/mL and no change in the temperature of the water.
  2. A 60-watt light bulb consumes 60 joules of energy each second that it is in use. A typical dishwasher might require about 1,500 joules per second to function. If one liter of sweat is vaporized over the course of an hour, how many joules per second are being absorbed by the water? Compare this value to the power requirements of a light bulb or a dishwasher.
  3. A large amount of energy is being used to vaporize sweat, but the temperature of the water molecules being vaporized is the same before and after evaporation occurs. Where is this energy going?
  4. Heat of vaporization can be used to estimate the strength of hydrogen bonding in water. At room temperature (25oC), the heat of vaporization of water is 44.0 kJ per mole of water molecules. A mole of liquid water contains about two moles of hydrogen bonds. How much energy (in kJ/mol) is required to break the hydrogen bonds in water at room temperature? How does this compare to the strength of the covalent O-H bonds within each water molecule (approximately 460 kJ/mol)?

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Image Attributions

  1. [1]^ Credit: Olaf Tausch; Source: http://commons.wikimedia.org/wiki/File:NaturTherme_Templin_Sauna_07.jpg; License: CC BY-NC 3.0
  2. [2]^ Credit: jeffrey montes; Source: http://www.flickr.com/photos/jm-photography/4728646934/; License: CC BY-NC 3.0

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