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.

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

- 70

this is the answer for you hope it