When you jog, most of the food energy you burn above your basalmetabolic rate (BMR) ends up as internal energy that would raise your bodytemperature if it were not eliminated. The evaporation of perspiration is theprimary mechanism for eliminating this energy. Determine the amount of wateryou lose to evaporation when running for 45 minutes at a rate that uses 450kcal/h above your BMR. (That amount is often considered to be the “maximumfat-burning” energy output.) The metabolism of 1 gram of fat generatesapproximately 9.0 kcal of energy and produces approximately 1 gram of water.(The hydrogen atoms in the fat molecule are transferred to oxygen to formwater.) What fraction of your need for water will be provided by fatmetabolism? (The latent heat of vaporization of water at room temperature is 2.5× 106 J/kg)
Question
When you jog, most of the food energy you burn above your basalmetabolic rate (BMR) ends up as internal energy that would raise your bodytemperature if it were not eliminated. The evaporation of perspiration is theprimary mechanism for eliminating this energy. Determine the amount of wateryou lose to evaporation when running for 45 minutes at a rate that uses 450kcal/h above your BMR. (That amount is often considered to be the “maximumfat-burning” energy output.) The metabolism of 1 gram of fat generatesapproximately 9.0 kcal of energy and produces approximately 1 gram of water.(The hydrogen atoms in the fat molecule are transferred to oxygen to formwater.) What fraction of your need for water will be provided by fatmetabolism? (The latent heat of vaporization of water at room temperature is 2.5× 106 J/kg)
Solution
First, let's calculate the total energy used in 45 minutes of running.
The rate of energy use is 450 kcal/h, so in 45 minutes (or 0.75 hours), the energy used is:
450 kcal/h * 0.75 h = 337.5 kcal
Next, we need to convert this energy from kcal to Joules, since the latent heat of vaporization is given in Joules.
1 kcal = 4184 J, so:
337.5 kcal * 4184 J/kcal = 1.41 * 10^6 J
The amount of water lost to evaporation can be calculated using the latent heat of vaporization, which is the amount of energy needed to change 1 kg of a substance from a liquid to a gas.
The latent heat of vaporization of water is 2.5 * 10^6 J/kg, so the mass of water evaporated is:
1.41 * 10^6 J / 2.5 * 10^6 J/kg = 0.564 kg
This is the amount of water lost to evaporation.
Next, let's calculate the amount of water produced by fat metabolism.
The energy from the metabolism of 1 gram of fat is 9.0 kcal, and this process produces 1 gram of water.
So, the mass of fat metabolized is:
337.5 kcal / 9.0 kcal/g = 37.5 g
Since 1 gram of fat produces 1 gram of water, this means that 37.5 g of water is produced by fat metabolism.
Finally, we can calculate the fraction of the need for water that is provided by fat metabolism:
37.5 g / 564 g = 0.067
So, about 6.7% of the water need is provided by fat metabolism.
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