Consider a 176lb male farmer is in 104 degree heat. On this day, the humidity is low enough such that all the sweat evaporates from the farmer's skin, enabling him to maintain a body temperature of 100 degrees. The farmer sweats 1 liter of water per hour and 60% of the latent heat needed to Evan=pirate this sweat is provided by farmer's body. If the farmer suddenly dehydrates and stops sweating, how long will it take for his body temperature to rise to the near-fatal 104 degrees.

t = 33 seconds

It is the amount of time when the temperature of the body will rise from 100 to fatal 104 in the absence of sweat.

Step-by-step explanation:

weight of the farmer = 176 lb

Sweating = 1 litre per hour

60% of above sweating is evaporated by the latent heat from farmers body.

Temperature outside = 104

Temperature of the Farmer = 100

Time to rise the temperature 4 degrees = ?

We need to use the specific heat formula to calculate the time to rise the temperature of the body of farmer.

Before, that let's convert weight of the farmer in Kg

W = 176 lb

1 lb = 0.453592 kg

176 lb = 79.8 kg

w = mg

g = 10

mass = m = 79.8/10

mass = 7.98 kg

Q = mc∆T

c = specific heat of water =  4,186

∆T  = Temperature Change = 104-100

∆T  = 4

Q = 7.98 x 4186 x 4

Q = 133617

t = Q/P

Power of Sun's Heat = 3.846×1026

t = 33 seconds

It is the amount of time when the temperature of the body will rise from 100 to fatal 104 in the absence of sweat.

here is the

mark brainliest pl

7

step-by-step explanation:

i'm assuming you need to figure out how many campers go into each tent.

i got the answer by:

24+4=28

28/4=7