A man consumes 3000 kilocalories of food in one day, converting most of it to thermal energy to maintain body temperature. If he loses half this energy by evaporating water through breathing and sweating, how many kilograms of water evaporate? For specific latent heat of vaporization of water, use 2430 kilojoules per kilogram.
Okay, so in this question, we’ve got a man who consumes 3000 kilocalories of food in one day. We’re told that he converts most of it to thermal energy to maintain body temperature. More specifically, we’re told he loses half this energy by evaporating water, through breathing and sweating of course, and we’re asked to find out how many kilograms of water evaporate. We are also told that the specific latent heat of vaporization of water is 2430 kilojoules per kilogram.
So the guy that we’re talking about consumes 3000 kilocalories of food every day. But we’re told that he loses half of this by evaporating water. In other words, half of this energy, half of 3000 kilocalories, goes towards evaporating water.
Now this energy will be provided as heat in order to evaporate the water. So we can label this energy in question as 𝑄 because we normally label heat as 𝑄. And this energy, of course, is 3000 divided by two because it’s half of the 3000 kilocalories he eats. This ends up being 1500 kilocalories.
Now we also need to understand what specific latent heat of vaporization actually means. The specific latent heat of vaporization, 𝐿, is defined as the heat required to change a kilogram of liquid to vapor. In other words, it’s the heat supplied per unit mass to convert liquid to vapor. Or in symbols, the latent heat of vaporization, 𝐿, is equal to 𝑄 over 𝑚 because it’s the heat supplied to a liquid divided by the mass of that liquid. That way, we find out the amount of heat supplied to each kilogram of that liquid. And of course, this heat goes towards converting the liquid to gas.
Now we’ve been given the value of 𝐿 for water. We know that as 2430 kilojoules per kilogram. So let’s look quickly at the units of 𝐿. As we expect, we have a unit of heat or energy divided by a unit of mass. Now this is all fine and dandy.
The problem comes in when we realize that the heat that we’ve been given is not in joules or kilojoules, but rather it’s in kilocalories, which means we need to either convert 𝑄 into kilojoules or convert 𝐿 into kilocalories per kilogram. Usually, it makes sense to convert quantities into their standard units, and the standard unit of energy or heat is the joule.
Now obviously, in our question, we don’t have anything in joules. We’ve only got it in either kilocalories or kilojoules, but a kilojoule is basically just 1000 joules, so let’s convert 𝑄 into kilojoules. At the end of the day, it doesn’t matter which of these we convert. We’ll get the same answer as long as we make sure we’re being consistent. If we’ve got 𝑄 in kilojoules, then we need 𝐿 in kilojoules per kilogram, or if we choose to go with kilocalories for 𝑄, then we need 𝐿 in kilocalories per kilogram. Either way, it doesn’t matter, so let’s just convert 𝑄.
The conversion factor is that one kilocalorie is equal to 4184 joules. Now we’re trying to convert 1500 kilocalories, so that ends up being 1500 times 4184 joules. Evaluating this gives us 6276000 joules. But remember, we’re trying to convert to kilojoules, not just to joules. The conversion is the following. One kilojoules is the same as 1000 joules, so we need to find the number of kilojoules that equals 6276000 joules. To do this, we just divide this number by 1000. And so we find 6276 kilojoules is the same as 6276000 joules, which means that we’ve officially converted 𝑄 into kilojoules. So we’ll replace 1500 kilocalories with 6276 kilojoules.
And now all that remains is for us to rearrange this equation to find out the mass of the water evaporated. Multiplying both sides of the equation by 𝑚 over 𝐿, we find that the mass of the water evaporated is equal to the heat supplied to the water divided by the specific latent heat of vaporization. Subbing in our values, we find that 𝑚 is equal to 6276 kilojoules divided by 2430 kilojoules per kilogram.
Looking at the units, the kilojoules cancel with these kilojoules, and essentially we’re left with one over kilogram in the denominator, as if we had kilogram in the numerator. Now evaluating the fraction gives us our final answer that the mass of water evaporated is 2.59 kilograms. That’s quite a lot of water to evaporate from a guy just breathing and sweating.