91Ó°ÊÓ

Consider a poor lost soul walking at 5 km/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 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 \(k'A{_s}{_k}{_i}{_n}(T{_a}{_i}{_r} - T{_s}{_k}{_i}{_n})\), where \(k'\) is 54 J/h \(\cdot\) C\(^\circ\) \(\cdot\) m\(^2\), the exposed skin area \(A{_s}{_k}{_i}{_n}\) is 1.5 m\(^2\), the air temperature \(T{_a}{_i}{_r} \)is 47\(^\circ\)C, and the skin temperature \(T{_s}{_k}{_i}{_n}\) is 36\(^\circ\)C; (iii) the skin absorbs radiant energy from the sun at a rate of 1400 W/m\(^2\); (iv) the skin absorbs radiant energy from the environment, which has temperature 47\(^\circ\)C. (a) Calculate the net rate (in watts) 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\(^\circ\)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\(^\circ\)C is \(2.42 \times 10{^6}\) J/kg.) (c) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only 0.45 m\(^2\). What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

Step by step solution

01

Calculate Metabolic Heat Generation

The metabolic reactions in the body generate 280 W of energy, all converted to heat. This value remains constant for calculations as it directly adds to the skin's thermal energy.

02

Calculate Heat by Convection

The heat transfer by convection is calculated using the formula: \[ Q_{conv} = k' \cdot A_{skin} \cdot (T_{air} - T_{skin}) \]Plugging in the values, \[ Q_{conv} = 54 \times 1.5 \times (47 - 36) = 891 \, \text{J/h} \]Convert this to watts: \[ Q_{conv} = \frac{891}{3600} \approx 247.5 \, \text{W} \]

03

Calculate Heat from Solar Radiation

The skin absorbs radiant energy from the sun at the rate of 1400 W/m² over the exposed skin area of 1.5 m². Therefore, \[ Q_{solar} = 1400 \times 1.5 = 2100 \, \text{W} \]

04

Calculate Heat from Environmental Radiation

Use the Stefan-Boltzmann law to calculate the absorbed energy from the environment:\[ Q_{env} = e \cdot \sigma \cdot A_{skin} \cdot (T_{air}^4 - T_{skin}^4) \]Assuming emissivity, \( e = 1 \), and \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4 \):Convert temperatures from Celsius to Kelvin: \\( T_{air} = 320 \, \text{K} \) and \( T_{skin} = 309 \, \text{K} \). The calculation becomes: \[ Q_{env} = 1 \times 5.67 \times 10^{-8} \times 1.5 \times (320^4 - 309^4) = 930.48 \, \text{W} \]

05

Calculate Net Heat Rate

Sum all the contributions (adding metabolic and radiant sources, subtracting convection as a cooling effect):\[ Q_{net} = 280 + 247.5 + 2100 + 930.48 = 3557.98 \, \text{W} \]

06

Determine Water Evaporation Rate for Constant Temperature

To maintain the skin temperature, the excess heat must be lost through evaporation:Let \( m \) be the mass of water evaporated in kg. The heat of vaporization is given as \( 2.42 \times 10^6 \, \text{J/kg} \). Solve for \( m \): \[ m = \frac{3557.98}{2.42 \times 10^6} \approx 0.00147 \, \text{kg/s} \] Convert this rate to L/h using 1 kg = 1 L: \[ \text{Rate} = 0.00147 \times 3600 \approx 5.29 \, \text{L/h} \]

07

Adjust Calculation for Clothing Protection

With clothing, the skin area is reduced to 0.45 m². Recaculate the heat contributions:- Convection: \[ Q_{conv} = 54 \times 0.45 \times (47 - 36) \approx 74.25 \, \text{W} \]- Solar: \[ Q_{solar} \approx 0 \, \text{W (assumed due to clothing)} \] - Environmental Radiation: \[ Q_{env} = 1 \times 5.67 \times 10^{-8} \times 0.45 \times (320^4 - 309^4) \approx 279.144 \, \text{W} \]Calculate new net heat rate:\[ Q_{net}' = 280 + 74.25 + 279.144 \approx 633.39 \, \text{W} \]Evaporation rate for clothing:\[ m' = \frac{633.39}{2.42 \times 10^6} \approx 0.000262 \, \text{kg/s} \] Convert to L/h:\[ 0.000262 \times 3600 \approx 0.94 \, \text{L/h} \]

08

Compare and Discuss Effectiveness of Clothing

The required evaporation rate with clothing (0.94 L/h) is significantly lower than without (5.29 L/h), demonstrating that traditional desert clothing effectively reduces the heat load on the body and helps conserve water.

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Key Concepts

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

Heat Transfer

Heat transfer is a fundamental concept in thermal physics, referring to the movement of thermal energy from one place to another due to temperature differences. In the context of the exercise, the lost soul's skin experiences heat transfer in several forms: metabolic heat generation, convective heat from the surrounding hot air, solar radiation absorption, and environmental radiation absorption. Each of these mechanisms contributes differently to the net heating effect on the individual's skin. Understanding heat transfer involves recognizing

• Conduction: Movement of heat through a solid medium.
• Convection: Transfer of heat through fluid motion.
• Radiation: Transfer of energy in the form of electromagnetic waves.

In this scenario, convection plays a critical role, with the temperature difference between the desert air and the skin driving a constant heat exchange. The formula used to calculate the heat delivered by convection is \[ Q_{conv} = k' imes A_{skin} imes (T_{air} - T_{skin}) \] where all the terms describe factors affecting the rate of heat transfer. The process allows us to determine how quickly heat is exchanged, shaping our understanding of energy dynamics in different environmental settings.

Metabolic Reactions

Metabolic reactions refer to the biochemical processes inside living organisms, where energy is produced and consumed to perform various activities within the cells. A byproduct of these reactions is heat, generated as the body metabolizes nutrients to keep functioning. In the exercise, metabolic reactions release 280 watts of heat continuously, contributing significantly to raising the individual's skin temperature. This internal heat production is estimated based on the physiological activity level and tissue composition of the person. Metabolic heat generation can be characterized as:

• Basal Metabolism: The minimal energy expenditure required to maintain vital functions at rest.
• Active Metabolism: Additional heat generated from activities like walking, as mentioned in the exercise.

In thermal physics, understanding the contribution of metabolic reactions to the overall heat balance is essential for calculating how organisms maintain thermal regulation under varying environmental conditions.

Radiation

Radiation is the transfer of heat through electromagnetic waves, and plays a crucial role in our exercise scenario. When energy is radiated from a source, it travels through space without needing a material medium. This property makes radiation uniquely capable of transferring heat in a vacuum or through clear air.For the poor lost soul exposed to the desert sun, radiation is a major source of heat. The sun emits energy across a spectrum that the skin absorbs efficiently, contributing a substantial 1400 W/m² over the exposed area. This is computed as:\[ Q_{solar} = ext{Radiant Flux} imes A_{skin} \] In addition, the environment also radiates energy, modeled by applying the Stefan-Boltzmann law, which describes how much energy is emitted or absorbed by a body with respect to its surface temperature:\[ Q_{env} = e \times \sigma \times A_{skin} \times (T_{air}^4 - T_{skin}^4) \] Understanding these processes can help optimize protective measures, such as wearing clothing with low emissivity, to minimize heat absorption from the environment.

Evaporation

Evaporation is a cooling mechanism vital for maintaining a constant body temperature, especially in hot environments. It involves the phase change of water from liquid to vapor, consuming energy in the process. This consumption of energy effectively removes heat from the skin, providing a cooling effect.In the exercise, to offset the internal and external heat sources, the lost soul's surplus heat of 3557.98 W should be counteracted through evaporation. The calculation for the rate of perspiration necessary to achieve thermal balance can be expressed as:\[ m = \frac{Q_{net}}{L_v} \] where \( Q_{net} \) is the net heat input and \( L_v \) is the latent heat of vaporization. Converting the mass of water evaporated per second to liters per hour helps to assess the practical implications in real-world scenarios.Evaporation rate reduction with protective clothing also underscores the traditional wisdom of desert dwellers, who leverage light-colored, loose-fitting attire to slow down thermal gain and conserve body water, showcasing the balance of modern science and age-old practices in maintaining comfort in extreme conditions.