The latent heat of vaporization of H2O at body temperature (37.0°C) is 2.41E+6 J/kg. To cool the body of a 76.7 kg jogger [average specific heat capacity = 3540 J/(kg*°C)] by 1.90°C, how many kilograms of water in the form of sweat have to be evaporated?

To calculate the amount of water that needs to be evaporated, we need to consider the heat gain and heat loss of the jogger's body.

First, let's calculate the heat gained by the jogger's body:
Heat gained = mass * specific heat capacity * change in temperature
Heat gained = (76.7 kg) * (3540 J/(kg*°C)) * (1.90°C)

Next, let's calculate the heat lost due to water evaporation:
Heat lost = mass of water * latent heat of vaporization

Since the heat gained and heat lost are equal (due to energy conservation), we can set the two equations equal to each other:

(76.7 kg) * (3540 J/(kg*°C)) * (1.90°C) = mass of water * (2.41E+6 J/kg)

Now, let's solve for the mass of water:

mass of water = ((76.7 kg) * (3540 J/(kg*°C)) * (1.90°C)) / (2.41E+6 J/kg)

mass of water ≈ 0.233 kg

Therefore, approximately 0.233 kg of water in the form of sweat needs to be evaporated to cool the jogger's body by 1.90°C.

To calculate the amount of water that needs to be evaporated, we can use the formula:

Q = mcΔT

Where:
Q is the amount of heat transferred,
m is the mass of the substance,
c is the specific heat capacity, and
ΔT is the change in temperature.

We will calculate the amount of heat transferred to the jogger's body and then divide that by the latent heat of vaporization of water to find the mass of water that needs to be evaporated.

First, calculate the amount of heat transferred to the jogger's body:

Q = mcΔT
Q = 76.7 kg * 3540 J/(kg*°C) * 1.90°C

Q = 514,431 J

Now, divide the amount of heat transferred by the latent heat of vaporization of water:

m = Q / latent heat of vaporization of water
m = 514,431 J / 2.41E+6 J/kg

m ≈ 0.213 kg

Therefore, approximately 0.213 kg of water in the form of sweat needs to be evaporated to cool the jogger's body.