1. Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. a. What mass of water must evaporate from the skin of a 70.0-kg man to cool his body 1.00 C°? The heat of vaporization of water at body temperature (37°C) is 2.42 X 10° J/kg·K. The specific heat of a typical human body is 3480 J/kg K b. (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can (355 cm).
To answer this question, we'll need to use the formulas for heat transfer and the relationship between mass, volume, and density of water.
a) To cool the body by 1.00°C, we'll need to calculate the heat transferred through the evaporation of water. The formula for heat transfer is:
Q = m * c * ΔT
Where:
- Q is the heat transferred
- m is the mass of water
- c is the specific heat of the human body
- ΔT is the change in temperature
Given values:
- c = 3480 J/kg K
- ΔT = 1.00°C = 1.00 K
We need to find the mass, so we rearrange the formula:
m = Q / (c * ΔT)
Now let's calculate:
m = Q / (3480 J/kg K * 1.00 K)
As for the heat of vaporization at body temperature, it is not necessary for this calculation, so we won't use it.
b) To determine the volume of water the man must drink to replenish the evaporated water, we'll need to compare it to the volume of a soft-drink can.
To calculate the volume, we need the density of water and the mass of water evaporated.
Given value:
- Density of water = 1 g/cm³ = 1000 kg/m³
Let's find the volume:
Volume = Mass / Density
Now, let's calculate:
Volume = m / (1000 kg/m³)
Finally, we can compare the volume of water to the volume of a soft-drink can, which is given as 355 cm³.
I hope this explanation helps you understand how to solve these types of problems.