91Ó°ÊÓ

During a period of relative inactivity, the average rate of transport of enthalpy by the metabolic and digestive waste products leaving the body minus the rate of enthalpy transport by the raw materials ingested and breathed into the body is approximately \(\Delta H=-300 \mathrm{kJ} / \mathrm{h}\). Heat is transferred from the body to its surroundings at a rate given by \(Q=h A\left(T_{\mathrm{s}}-T_{0}\right)\) where \(A\) is the body surface area (roughly \(1.8 \mathrm{m}^{2}\) for an adult), \(T_{\mathrm{s}}\) is the skin temperature (normally \(\left.34.2^{\circ} \mathrm{C}\right), T_{0}\) is the temperature of the body surroundings, and \(h\) is a heat transfer coefficient. Typical values of \(h\) for the human body are \(^{5}\) \(h=8 \mathrm{kJ} /\left(\mathrm{m}^{2} \cdot \mathrm{h} \cdot^{\circ} \mathrm{C}\right) \quad\) (fully clothed, slight breeze blowing) \(h=64 \mathrm{kJ} /\left(\mathrm{m}^{2} \cdot \mathrm{h} \cdot^{\circ} \mathrm{C}\right) \quad\) (nude, immersed in water) (a) Consider the human body as a continuous system at steady state. Write an energy balance on the body, making all appropriate simplifications and substitutions. (b) Calculate the surrounding temperature for which the energy balance is satisfied (i.e., at which a person would feel neither hot nor cold) for a clothed person and for a nude person immersed in water. (c) At a family party, an elderly relative calls out to you in a loud voice, "Hey, you're an engineer, so you know everything. Explain why I'm comfortable when the room temperature is seventy degrees, but if I get into seventy-degree water in a bathtub, I'm freezing." He stops with a smug expression on his face and along with everyone else within earshot waits for your response. What would it be?

Step by step solution

01

Equation Formulation

For part (a), let's formulate the energy balance equation. At steady-state, energy balance equation is given by the equation \(\dot{Q}_{in} - \dot{Q}_{out} - \Delta \dot{H} = 0\). Since in the given problem, there is no heat given to the body, the equation simplifies to \(-\dot{Q}_{out} - \Delta \dot{H} = 0\), where the heat transfer from the body to the surrounding is given by the formula \(\dot{Q}_{out} = h A (T_s - T_0)\). Thus, replacing \(\dot{Q}_{out}\) in the energy balance equation yields \(-h A (T_s - T_0) - \Delta \dot{H} = 0\). Given that, \(T_s = 34.2^{\circ}\) C, \(A = 1.8 m^2\), \(\Delta \dot{H} = -300 kJ/h\), the equation can be rearranged for \(T_0 = T_s - \frac{\Delta \dot{H} - A}{hA}\) C.

02

Energy Balance Calculation

We continue to part (b) where we need to calculate for \(T_0\) for a clothed person and for a nude person immersed in water. For a fully clothed person, \(T_0 = 34.2 - \frac{-300}{8*1.8} = 34.2 + 20.83 = 55.03^{\circ}\) C. For a person nude in water, \(T_0 = 34.2 - \frac{-300}{64*1.8} = 34.2 + 2.60 = 36.8^{\circ}\) C. Thus, the person would feel comfortable at 55.03 \(^{\circ}\) C if fully clothed and at 36.8 \(^{\circ}\) C if nude in water.

03

Understanding the Feeling of Cold

Finally, moving to part (c), it's important to note that the rate at which heat is transferred from the body depends on the value of \(h\), the heat transfer coefficient. For an elderly family member feeling colder in water at 70 \(^{\circ}\) F than in air at the same temperature, this can be explained by the fact that the heat transfer coefficient in water is much greater than in air (as given as 64 \(kJ/m^2 h \(^{\circ}\) C\) and 8 \(kJ/m^2 h \(^{\circ}\) C\) respectively). This means heat is lost from the body to the water at a much faster rate than it is to air even when the temperatures are the same. Thus, water will feel colder than air at the same temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer

The phenomenon of heat transfer is crucial in understanding how heat moves from one medium to another. In the context of the human body, heat transfer happens when body heat is exchanged with the surroundings.
This process is measured using the heat transfer coefficient, denoted as \( h \), which reflects how effectively heat is transferred. The rate of heat transfer \( Q \) is calculated with the formula:

• \( Q = hA(T_s - T_0) \)

This equation shows that heat transfer depends on:

• The surface area \( A \) of the body, roughly \( 1.8 m^2 \).
• The temperature difference \( T_s - T_0 \) between the skin and surroundings.

Typical \( h \) values differ significantly depending on conditions:
- 8 \( kJ/(m^2 \cdot h \cdot ^\circ C) \) when wearing clothes with a slight breeze, and
- 64 \( kJ/(m^2 \cdot h \cdot ^\circ C) \) when nude and immersed in water.
These values highlight how different environments impact the body's heat loss rate. Under standing conditions, heat transfer ensures temperature regulation.

Metabolic Processes

Metabolic processes involve chemical reactions within the body that help maintain vital functions, such as heart rate and digestion. During these processes, enthalpy changes occur, which are a form of energy balance within the body. These reactions result in a change in enthalpy, known as \( \Delta H \).
In periods of rest, the net metabolic energy change is negative, such as \( \Delta H = -300 \) kJ/hr, because the body is releasing more energy than it consumes. This energy needs to be balanced by the body's heat transfer mechanisms to maintain a constant internal temperature.
Without these processes efficiently managing energy transformations, the body's energy balance would be disrupted, affecting overall health. This careful orchestration allows us to feel stable under varying external conditions.

Steady State Analysis

Steady state analysis is pertinent to understanding systems where variables remain consistent over time. In this context, the human body can be analyzed in steady state when it maintains a constant internal temperature despite external fluctuations.
In a steady state, the energy balance equation is simplified:

• \( \dot{Q}_{in} - \dot{Q}_{out} - \Delta \dot{H} = 0 \)

For the human body, simplifying further gives:

• \( -\dot{Q}_{out} - \Delta \dot{H} = 0 \)

This means that the heat lost to the environment, \( \dot{Q}_{out} \), must equal the internal energy change \( \Delta \dot{H} \) to maintain temperature.
This understanding helps explain how the human body adapts to different temperatures by modifying the heat transfer rate to remain in equilibrium. It also elucidates why certain ambient temperatures are more comfortable: the rate of heat loss is balanced with metabolic energy release, ensuring the body neither overheats nor cools excessively.