How Much Sweat Evaporates to Cool a 80kg Man by 1°C?

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SUMMARY

The discussion focuses on calculating the mass of sweat required to cool an 80 kg man by 1°C through evaporation. The heat of evaporation of water at body temperature (37°C) is established as 2400 J/g. Using the formula Q = Lm, where Q is the heat lost, L is the latent heat of evaporation, and m is the mass of sweat, participants aim to derive the necessary mass of sweat for the temperature decrease. The solution involves applying the equation c = ∆Q ÷ m∆T to find the mass of water evaporated.

PREREQUISITES
  • Understanding of thermodynamics, specifically heat transfer.
  • Familiarity with the concept of latent heat of evaporation.
  • Basic algebra for manipulating equations.
  • Knowledge of specific heat capacity and its application in temperature change calculations.
NEXT STEPS
  • Study the principles of heat transfer in thermodynamics.
  • Learn about latent heat and its significance in phase changes.
  • Explore specific heat capacity calculations and their applications.
  • Practice solving problems involving mass, heat, and temperature changes using relevant equations.
USEFUL FOR

Students in physics or engineering, educators teaching thermodynamics, and anyone interested in the physiological aspects of human cooling mechanisms through sweat evaporation.

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Homework Statement



How large is the mass of sweat (water) evaporated by a man with 80 kg body weight assuming that his body
temperature decreased due to the evaporation by 1 °C? (Assume, that the human body consists mostly of water
and the heat of evaporation of water at body temperature (37 °C) is around 2400 J/g) (14 dkg)

Homework Equations



Q = Lm

c = ∆Q ÷ m∆T

The Attempt at a Solution



Many, so far. But I have no idea which variable for mass to use in which equation. I have almost no time left and need help quickly! The best would be a step-through-step.
 
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