Question

Suppose a person metabolizes \(2000 .\) kcal/day. a) With a core body temperature of \(37.0^{\circ} \mathrm{C}\) and an ambient temperature of \(20.0^{\circ} \mathrm{C}\), what is the maximum (Carnot) efficiency with which the person can perform work? b) If the person could work with that efficiency, at what rate, in watts, would they have to shed waste heat to the surroundings? c) With a skin area of \(1.50 \mathrm{~m}^{2}\), a skin temperature of \(27.0^{\circ} \mathrm{C}\) and an effective emissivity of \(e=0.600,\) at what net rate does this person radiate heat to the \(20.0^{\circ} \mathrm{C}\) surroundings? d) The rest of the waste heat must be removed by evaporating water, either as perspiration or from the lungs At body temperature, the latent heat of vaporization of water is \(575 \mathrm{cal} / \mathrm{g}\). At what rate, in grams per hour, does this person lose water? e) Estimate the rate at which the person gains entropy. Assume that all the required evaporation of water takes place in the lungs, at the core body temperature of \(37.0^{\circ} \mathrm{C}\).

Short answer

Answer: The person needs to shed waste heat at a rate of 91.89 watts.

Step-by-step solution

Find the temperature difference

Calculate the difference between the core body temperature \(T_{core}=37.0^{\circ} \mathrm{C}\) and the ambient temperature \(T_{ambient}=20.0^{\circ} \mathrm{C}\): \(\Delta T = T_{core} - T_{ambient} = 37.0 - 20.0 = 17.0^{\circ} \mathrm{C}\)

Convert to Kelvin

Convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature: \(T_{core\_Kelvin} = T_{core} + 273.15 = 37.0 + 273.15 = 310.15\,\text{K}\) \(T_{ambient\_Kelvin} = T_{ambient} + 273.15 = 20.0 + 273.15 = 293.15\,\text{K}\)

Calculate the maximum efficiency

For a Carnot engine, the efficiency is given by: \(\eta = 1 - \frac{T_{cold}}{T_{hot}}\) Calculate the maximum efficiency using the given temperatures: \(\eta = 1 - \frac{T_{ambient\_Kelvin}}{T_{core\_Kelvin}} = 1 - \frac{293.15}{310.15} \approx 0.0548\) The maximum Carnot efficiency is about 5.48%. #b) The rate to shed waste heat in watts#

Calculate energy expenditure

The person metabolizes 2000 kcal/day. Convert this to joules and find the energy per second (power in watts) using the conversion \(1\,\text{kcal} = 4184\,\text{J}\): \(P_{total} = \frac{2000\,\text{kcal} \cdot 4184\,\text{J/kcal}}{24 \cdot 3600\,\text{s/day}} \approx 97.22\,\text{W}\)

Calculate useful work

Find the amount of energy converted into useful work by multiplying the total power by the efficiency: \(P_{work} = \eta \cdot P_{total} \approx 0.0548 \cdot 97.22 \approx 5.329\,\text{W}\)

Calculate waste heat

Subtract the useful work from the total power to find the rate at which waste heat is shed: \(P_{waste} = P_{total} - P_{work} \approx 97.22 - 5.329 \approx 91.89\,\text{W}\) The person needs to shed waste heat at a rate of 91.89 watts. #c) The net rate of heat radiation#

Use Stefan-Boltzmann Law

The net rate of heat radiation is given by: \(P_{rad} = e A \sigma \left(T_{skin}^4 - T_{ambient}^4\right)\) where \(e\) is the effective emissivity, \(A\) is the skin area, \(\sigma\) is the Stefan-Boltzmann constant (5.67 x 10-8 W/m²·K⁴), and \(T_{skin}\) and \(T_{ambient}\) are the skin and ambient temperatures in Kelvin, respectively.

Calculate the net rate of heat radiation

Plug in the given values and solve for \(P_{rad}\): \(P_{rad} = 0.600 \cdot 1.50\,\text{m}^2 \cdot 5.67 \times 10^{-8}\,\text{W/m}^2\text{K}^4 \cdot \left((300.15\,\text{K})^4 - (293.15\,\text{K})^4\right) \approx 40.87\,\text{W}\) The net rate of heat radiation is about 40.87 watts. #d) Rate at which person loses water#

Calculate the rate of heat removal by evaporation

Subtract the net rate of heat radiation from the rate of waste heat shedding to find the rate of heat removal by evaporation: \(P_{evap} = P_{waste} - P_{rad} \approx 91.89 - 40.87 \approx 51.02\,\text{W}\)

Convert heat rate to water mass rate

Using the latent heat of vaporization of water at body temperature, 575 cal/g, we can convert the power rate to a mass rate: \(\frac{\text{d}m}{\text{d}t} = \frac{P_{evap}}{L_v} = \frac{51.02\,\text{W}}{575\,\text{cal/g} \cdot 4.184\,\text{J/cal}} \approx 0.0250\,\text{g/s}\) Convert the mass rate to grams per hour: \(\frac{\text{d}m}{\text{d}t} = 0.0250\,\text{g/s} \cdot \frac{3600\,\text{s}}{\text{hour}} \approx 90.0 \, \text{g/hour}\) The person loses water at a rate of 90.0 grams per hour. #e) Estimate the rate of entropy gain#

Calculate the heat flow

Assuming all the evaporation takes place in the lungs at the body temperature, calculate the heat flow due to evaporation: \(Q = P_{evap} = 51.02\,\text{W}\)

Calculate the rate of entropy gain

The rate of entropy gain is given by: \(\frac{\text{d}S}{\text{d}t} = \frac{Q}{T_{core\_Kelvin}} = \frac{51.02\,\text{W}}{310.15\,\text{K}} \approx 0.164\,\text{W/K}\) The person gains entropy at a rate of approximately 0.164 watts per kelvin.

Key concepts

Metabolism in Physics

In the context of physics, metabolism refers to the process by which living organisms convert food into energy. It's a complex series of chemical reactions that take place in the cells, resulting in the release of energy, which is measured in calories or joules. Metabolism involves both the breaking down of molecules to obtain energy (catabolism) and the synthesis of all compounds needed by the cells (anabolism).

The concept of metabolism is important in thermodynamics because it illustrates how energy is transformed within living organisms and how this energy can do work, such as the human body performing physical tasks. Furthermore, considerations of efficiency come into play—this is where the Carnot efficiency provides a theoretical limit to the efficiency of energy conversion processes, even in biological systems like humans.

Stefan-Boltzmann Law

The Stefan-Boltzmann law is a fundamental principle in thermodynamics that explains how the power radiated by a black body (an idealized physical body that absorbs all incident electromagnetic radiation) is related to its temperature. Specifically, the law states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body’s absolute temperature.

Mathematically, it is expressed as: \[ P = e \sigma A (T^4) \] where \( P \) is the power radiated per unit area, \( e \) is the emissivity of a material (a measure of how effectively it radiates energy), \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the area, and \( T \) is the absolute temperature in kelvins.

This law is crucial when calculating the radiant heat loss from a human body and helps in understanding how variations in temperature and surface area can affect this radiative cooling process, as exemplified in our original exercise.

Entropy in Thermodynamics

Entropy is a measure of the disorder or randomness in a thermodynamic system and is a core concept in the second law of thermodynamics. It is often associated with the amount of 'waste' energy in a system that cannot be used to do work. The higher the entropy, the greater the disorder and the less available energy to do work.

In thermodynamics, the change in entropy \( (\Delta S) \) can be calculated using the formula: \[ \Delta S = \frac{Q}{T} \] where \( Q \) is the heat transfer into or out of the system, and \( T \) is the absolute temperature at which the transfer takes place.

For living organisms, entropy must be managed through metabolic processes. In the example provided, the human body generates entropy during metabolic processes, such as the evaporation of water from the lungs. The rate at which the body gains entropy has implications for its thermal regulation and energy efficiency. This ties back into the concept of metabolism as the body must expend energy to manage entropy and maintain order within its systems.