∙Bicycling on a warm day. If the air temperature is the same...

$\bullet$ Bicycling on a warm day. If the air temperature is the same as the temperature of your skin (about $30^{\circ} \mathrm{C}$ ), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 70 kg person's body produces energy at a rate of about 500 $\mathrm{W}$ due to metabolism, 80$\%$ of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is $2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}$ (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750 $\mathrm{mL}$ bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is 1.0 kg.)

$\bullet$ Bicycling on a warm day. If the air temperature is the same as the temperature of your skin (about $30^{\circ} \mathrm{C}$ ), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 70 kg person's body produces energy at a rate of about 500 $\mathrm{W}$ due to metabolism, 80$\%$ of which is converted
to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is $2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}$ (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750 $\mathrm{mL}$ bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is 1.0 kg.)

Key Concepts

Unit Conversion and Dimensional Analysis

This fundamental technique involves converting physical quantities between different units and ensuring dimensional consistency in calculations. It is essential for accurately determining energy, mass, and volume relationships in physical and physiological problems.

Evaporative Cooling

Evaporative cooling is the mechanism by which the body dissipates excess heat through the evaporation of sweat. This process is vital for maintaining a stable body temperature, especially when environmental conditions limit other forms of heat transfer.

Metabolic Energy Production

This concept refers to the process by which the body converts chemical energy into work and heat during physical activity. Understanding the rate at which energy is produced, usually measured in watts, is crucial when analyzing the energy output of the body and how that energy contributes to heat generation.

Heat Conversion Efficiency

In many biological processes, not all the energy produced is used for mechanical work; a significant proportion is converted to heat. This efficiency factor is important for determining the total amount of heat generated by metabolic processes, particularly during exercise in warm conditions.

Heat of Vaporization

This is a key thermodynamic property that specifies the amount of energy needed to convert a unit mass of a liquid into vapor at a given temperature. In the context of human physiology, it explains how the evaporation of sweat removes heat from the body by absorbing a substantial amount of energy.

Transcript

00:01 In this problem, we're going to talk about the latent heat.

00:04 So what we need to remember is that the latent heat is the energy necessary to transfer to a system or from a system in order to change its phase of matter.

00:18 And q is equal to the latent heat constant l times the mass of the object.

00:25 And what we have in our problem is a cyclist who produces energy at a rate at a power of 500 watts.

00:39 And about 80 % of this energy is produced in the form of the heat.

00:47 And in question a, we have to calculate how much mass of water a person must evaporate in order to lose all this heat that was produced.

01:02 So first, notice that the power that is produced in the form of heat is actually 80 % of 500, and that is 400 watts.

01:14 And in one hour, because we want to calculate the mass of water that must be evaporated in one hour, in one hour, the total energy is 400 watts times 360.

01:30 3 ,600 seconds.

01:34 So this is equal to 1 .44 times 10 to the 6th joules.

01:40 In one hour, the person must, the person produces in heat an energy that is equal to 1 .44 times 10 to the 6th joules.

01:53 The mass of water times the latent heat must be equal to this energy, e, because this so that the water, the evaporated water, can take away the heat from the person's body.

02:10 So if we want to calculate m, this is equal to the energy divided by l.

02:15 So that's 1 .44 times 10 to the 6 joules divided by the latent heat...

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