The latent heat of vaporizaiton of H2O at body temperature (37.0°C) is 2.42 106 J/kg. To cool the body of a 89 kg jogger (average specific heat capacity = 3500 J/(kg·C°)) by 1.4 C°, how many kilograms of water in the form of sweat have to be evaporated?
masssweat*Lv=massbody*c*deltaTemp
(89)(3500)(1.4)=M(2.42E^6)
436100=M(2.42E^6)
M=.180
To find the kilograms of water in the form of sweat that need to be evaporated, we need to calculate the amount of energy required to cool the jogger's body and then divide it by the latent heat of vaporization of water.
First, let's calculate the amount of energy required to cool the jogger's body:
Energy = mass * specific heat capacity * temperature change
mass = 89 kg
specific heat capacity = 3500 J/(kg·C°)
temperature change = 1.4°C
Energy = 89 kg * 3500 J/(kg·C°) * 1.4°C
Next, we divide the energy by the latent heat of vaporization of water:
Kilograms of water = Energy / latent heat of vaporization of water
latent heat of vaporization of water = 2.42 * 10^6 J/kg
Kilograms of water = (89 kg * 3500 J/(kg·C°) * 1.4°C) / (2.42 * 10^6 J/kg)
Now we can calculate the value:
Kilograms of water = (89 * 3500 * 1.4) / (2.42 * 10^6)
After solving the above expression, we can find the number of kilograms of water that need to be evaporated to cool the jogger's body.