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

AI Thread Summary
To determine the mass of sweat evaporated to cool an 80 kg man by 1°C, the heat loss required can be calculated using the formula Q = mcΔT, where m is mass, c is the specific heat capacity of the body, and ΔT is the temperature change. Given that the heat of evaporation of water is approximately 2400 J/g, the total energy loss needed for a 1°C decrease can be calculated. The discussion highlights confusion regarding which variables to use in the equations and requests a step-by-step solution due to time constraints. Understanding the relationship between heat loss and mass of evaporated sweat is crucial for solving the problem. A clear explanation of the equations and their application is necessary for a successful resolution.
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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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