Problem 6.

Even if a man shows no visible perspiration he still evaporates about 500 grams of water per day from his lungs. How many calories of heat are removed by this evaporation? What is the rate of heat loss in watts due to this process?

Solution:

First we calculate the energy (heat) required to evaporate the water in SI units.

The SI unit of mass is kg. We need to convert 500 g in kg: .

The heat required to vaporize the water is determined by the specific latent heat of vaporization:

The specific latent heat of vaporization (in SI units) is 2260000 J/kg. Then

Now we can convert the energy (expressed in joules) into calories. We need to use the relation between joules and calories:

1 joule = 0.24 calories

Then

Now we can find the rate of heat loss (in watts) – the power. The rate of heat loss can be calculated as the heat loss per unit time. The watt is a SI unit. It means that we need to calculate the heat (energy) loss (in joule) per 1 second (SI unit of time).

We know that the energy is the energy loss per one day. To find the energy loss per 1 second we need to find the number of seconds in one day: