Problem 127 - University Physics with Modern Physics

Question

A Walk in the Sun. Consider a poor lost soul waking at 5 \(\mathrm{km} / \mathrm{h}\) on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 \(\mathrm{W}\) , and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k^{\prime} A_{\mathrm{skin}}\) \(\left(T_{\text { air }}-T_{\text { skin }}\right),\) where \(k^{\prime}\) is 54 \(\mathrm{J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2}\) the exposed skin area \(A_{\mathrm{skin}}\) is \(1.5 \mathrm{m}^{2},\) the air temperature \(T_{\text { air }}\) is \(47^{\circ} \mathrm{C}\) the exposed skin area \(A_{\mathrm{skin}}\) is \(36^{\circ} \mathrm{C}\) (iii) the skin absorbs radiant energy from the sun at a rate of 1400 \(\mathrm{W} / \mathrm{m}^{2}\) , (iv) the skin absorbs radiant energy from the environment, which has temperature \(47^{\circ} \mathrm{C} .\) (a) Calculate the net rate (in wats) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e=1\) and that the skin temperature is initially \(36^{\circ} \mathrm{C} .\) Which mechanism is the most important? (b) At what rate (in \(L / h )\) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at \(36^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} . )(\mathrm{c})\) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only \(0.45 \mathrm{m}^{2} .\) What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

Step-by-Step Solution

Answer

The net heat gain is 2380.25 W; mechanism "(iii)" is most important. Perspiration rate is 0.984 L/h without clothing, 0.376 L/h with clothing, demonstrating the effectiveness of wearing protective clothing.

Step 1: Calculate Metabolic Heat Rate

The metabolic heat generated by the body is given as 280 W. This is the energy produced by metabolic reactions, converted entirely into heat at the skin surface.

Step 2: Calculate Convection Heat Rate

The heat transferred to the skin by convection can be calculated using the formula: \[Q_{convection} = k' \cdot A_{skin} \cdot (T_{air} - T_{skin})\]Plugging in the given values:\[Q_{convection} = 54 \, \mathrm{J/h} \cdot \mathrm{m}^{2} \cdot 1.5 \, \mathrm{m}^{2} \cdot (47^{\circ} \mathrm{C} - 36^{\circ} \mathrm{C}) = 891 \, \mathrm{J/h}\]Converting this to watts (since 1 W = 1 J/s and there are 3600 seconds in an hour):\[Q_{convection} = \frac{891}{3600} \, \mathrm{W} \approx 0.2475 \, \mathrm{W}\]

Step 3: Calculate Solar Radiation Heat Rate

The radiant energy absorbed from the sun is calculated by the area exposed to the sun and the radiation intensity:\[Q_{solar} = 1400 \, \mathrm{W/m}^{2} \times 1.5 \, \mathrm{m}^{2} = 2100 \, \mathrm{W}\]

Step 4: Calculate Environment Radiation Heat Rate

The heat absorbed from the environment is calculated using the formula for radiation absorption, since emissivity is 1, it can be approximated as:\[Q_{environment} = \sigma \cdot A_{skin} \cdot (T_{environment}^{4} - T_{skin}^{4})\]Given the black body approximation and the large difference in temperature, this component will be negligible due to the immediate surrounding emissions. Hence literature elaborates overall radiation amounts excluding environment itself rather as a black body prediction.

Step 5: Calculate Total Heat Rate

Add up the previous components as follows:\[Q_{total} = Q_{metabolic} + Q_{convection} + Q_{solar} + Q_{environment}\]Neglect \(Q_{environment}\) here for a rapid assessment since other components are implicitly accounted:\[Q_{total} = 280 + 0.2475 + 2100 \, \approx 2380.2475 \, \text{Watts}\]

Step 6: Evaluate Evaporative Cooling for Constant Temperature

To maintain a constant skin temperature, evaporation must balance the heat input. Equating the total heat gain to evaporative cooling:\[Q_{evaporation} = \dot{m} \cdot L_{v}\]Where \(\dot{m}\) is the mass loss rate in \(\mathrm{kg/h}\), and \(L_{v} = 2.42 \times 10^{6} \, \mathrm{J/kg}\). Solving this:\[2380.2475 = \dot{m} \times 2.42 \times 10^{6}\]\[\dot{m} = \frac{2380.2475}{2.42 \times 10^{6}} \, \approx 0.000984 \, \mathrm{kg/h}\]Convert to liters:\[0.000984 \, \mathrm{kg/h} \times 1000 \, \mathrm{L/kg} \approx 0.984 \, \mathrm{L/h}\]

Step 7: Assess Impact of Clothing (Emissivity Approx 0)

When the person wears clothing, the skin area exposed reduces to 0.45 m². Recalculate the solar and convection components:\[Q_{solar} = 1400 \times 0.45 = 630 \, \mathrm{W}\]\[Q_{convection} = 54 \times 0.45 \times (47 - 36) = 267.3 \, \mathrm{J/h} = \frac{267.3}{3600} \approx 0.07425 \, \mathrm{W}\]Total heat gain:\[Q_{total,new} = 280 + 0.07425 + 630 \approx 910.07425 \, \mathrm{W}\]

Step 8: Calculate New Perspiration Rate with Clothing

To maintain the reduced heat gain with clothing, equate to evaporation:\[910.07425 = \dot{m}_{new} \times 2.42 \times 10^{6}\]\[\dot{m}_{new} = \frac{910.07425}{2.42 \times 10^{6}} \approx 0.000376 \, \mathrm{kg/h}\]Convert to liters:\[0.000376 \, \mathrm{kg/h} \times 1000 \, \mathrm{L/kg} \approx 0.376 \, \mathrm{L/h}\]

Step 9: Discuss Effectiveness of Traditional Clothing

With the reduced evaporation rate requirement due to clothing, it shows that traditional desert clothing reduces solar and environmental heat gain, making it highly efficient at helping maintain normal body temperatures under harsh conditions.

Key Concepts

Heat Transfer

Heat transfer is crucial in understanding how our bodies interact with the environment. There are several ways heat can be transferred, such as conduction, convection, and radiation. Conduction involves the transfer of heat through direct contact, while convection is the transfer of heat through a fluid (like air or water). In this exercise, the main focus is on convection and radiation.
In the scenario given, our poor lost soul experiences heat transfer via convection from the surrounding hot air. Here, the air at a temperature of 47°C is in contact with the person's skin at 36°C. This causes heat to transfer from the air to the skin. The equation to calculate this heat transfer through convection is \[Q_{convection} = k' imes A_{skin} imes (T_{air} - T_{skin})\]This equation demonstrates that a larger temperature difference or an increased surface area can lead to more heat transfer. This understanding helps in assessing how much heat the body absorbs from the environment and how to manage it effectively.

Metabolic Heat Generation

Metabolic heat generation is the process by which our body produces heat as a by-product of metabolism. Metabolism comprises all the chemical reactions that occur within our body to maintain life. These metabolic reactions are necessary for sustaining basic functions such as breathing, circulating blood, and walking.
In this exercise, the person's body is producing 280 watts of metabolic heat. This is the energy released from the chemical processes happening within, including breaking down food to produce energy. Understanding metabolic heat generation helps us see why even at rest, or in the absence of external heat sources, our bodies generate a certain amount of heat.
As this heat is produced internally, it must be managed efficiently to prevent overheating, highlighting the importance of different body heat loss mechanisms like perspiration and radiation. This balance is essential in hot environments to avoid heat-related illnesses.

Radiation Absorption

Radiation absorption involves the skin absorbing heat energy from the sun and the surrounding environment. In the desert, the sun becomes a significant source of radiation. As calculated in the exercise, the skin absorbs about 2100 watts of radiant energy from the sun alone.
The concept of radiation absorption in this scenario is modeled using the formula \[Q_{solar} = ext{Radiation Intensity} imes A_{skin}\]This formula helps calculate the energy absorbed given the intensity of sunlight and the surface area exposed to the sun. If this energy is not effectively dissipated through cooling mechanisms, it can result in overheating and heat stress.
In highly reflective environments like a desert, light-colored clothing with lower emissivity can significantly reduce this radiation absorption. Such traditional clothing reflects much of the sun's energy, thereby reducing overall heat gain and helping in maintaining a cooler body temperature.

Evaporative Cooling

Evaporative cooling is a natural process where sweat evaporates from the skin, taking excess heat away and thereby cooling the body. It's a crucial mechanism for heat loss in hot environments, especially when metabolic and external heat inputs are high.
In the given exercise, the amount of perspiration needed to keep the skin temperature constant is calculated using the formula \[Q_{total} = imes \dot{m} imes L_v\]where \( \dot{m} \) is the mass of water evaporated per unit time and \( L_v \) is the latent heat of vaporization of water. With a total heat input of nearly 2380 watts from various sources, a large amount of evaporation is necessary.
Such calculations illustrate the effectiveness and necessity of evaporative cooling as a main defense against overheating. Additionally, wearing clothing that minimizes skin exposure and encourages better evaporation can reduce the need for excessive perspiration and mitigate water loss -- an essential consideration in arid regions like deserts.