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.
FFFUUU
Messages
2
Reaction score
0

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.
 
Physics news on Phys.org
Anyone?
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Question about Phase Change & Latent Heat
Replies
13
Views
3K
How Does Latent Heat Affect Sweat Evaporation in Cooling the Body?
Replies
3
Views
5K
How much ice to cool down this water?
Replies
17
Views
5K
B How does sweating cool a person down?
Replies
9
Views
15K
Why Does Liquid Cool During Evaporation?
Replies
3
Views
6K
How much water evaporates per minute from a copper pot on a stove?
Replies
4
Views
1K
Calculate the Mass of Ice cubes needed to cool a soft drink
Replies
10
Views
3K
Thermodynamics - Hypothermia - Heat loss
Replies
12
Views
2K
How Much Water Can a 1KW Carnot Engine Freeze in 5 Minutes?
Replies
3
Views
1K
Fraction within a fraction -- Units
Replies
3
Views
3K

Hot Threads

Recent Insights

Back
Top