Problem 127 - University Physics with Modern Physics

Question

A Walk in the Sun. Consider a poor lost soul walking 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 con- verted 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 \(k^{\prime} A_{\text { skin }}\left(T_{\text { air }}-T_{\text { skin }}\right),\) where \(k^{\prime}\) is \(54 \mathrm{J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2},\) the exposed skin area \(A_{\text { skin }}\) is \(1.5 \mathrm{m}^{2},\) the air temperature \(T_{\mathrm{air}}\) is \(47^{\circ} \mathrm{C},\) and the skin temperature \(T_{\text { skin }}\) 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 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 \(36^{\circ} \mathrm{C}\) . Which mechanism is the most important? (b) At what rate (in \(\mathrm{L} / \mathrm{h} )\) 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 \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\)) (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

(a) 3461.75 W, radiant heat from the sun is most important. (b) 5.15 L/h. (c) 1.84 L/h; clothing reduces perspiration need.

Step 1: Calculate Metabolic Heat

The metabolic heat generated by the body is given as 280 W. This value is straightforward as it is an input to the problem and does not require further calculation.

Step 2: Heat from Convection

To calculate the heat flow due to convection, use the formula: \[ Q_{\text{convection}} = k^{\prime} A_{\text{skin}} (T_{\text{air}} - T_{\text{skin}}) \] Given: - \( k^{\prime} = 54 \ \mathrm{J/h \cdot C^{\degree} \cdot m^{2}} \),- \( A_{\text{skin}} = 1.5 \ \mathrm{m^2} \),- \( T_{\text{air}} = 47^{\degree} \mathrm{C} \),- \( T_{\text{skin}} = 36^{\degree} \mathrm{C} \).Substitute these values into the equation to find \( Q_{\text{convection}} \). First, calculate the temperature difference: \[ 47 - 36 = 11 \ \mathrm{C^{\degree}} \] Then, calculate the heat transfer: \[ Q_{\text{convection}} = 54 \times 1.5 \times 11 = 891 \ \mathrm{J/h} \]Convert to watts by dividing by 3600 (since 1 W = 1 J/s): \[ Q_{\text{convection}} = \frac{891}{3600} \approx 0.2475 \ \mathrm{W} \]

Step 3: Absorbed Radiant Heat from the Sun

Calculate the absorbed radiant heat from the sun using the expression:\[ Q_{\text{sun}} = 1400 \times A_{\text{skin}} \] Substitute the value for the exposed skin area \( A_{\text{skin}} = 1.5 \mathrm{m^2} \):\[ Q_{\text{sun}} = 1400 \times 1.5 = 2100 \ \mathrm{W} \]

Step 4: Absorbed Radiant Heat from Environment

For environmental radiant heat, consider the effect of the air temperature (\(47^{\degree} \mathrm{C} \)) as the effective radiation temperature. However, simplifications or given assumptions are necessary (i.e. consider full absorption by emissive heat equation or neglect simplifications without thermal data).Using:\[ Q_{\text{environment}} = e \cdot \sigma A (T_{\text{air}} + 273)^{4} \] Substitute: - \( e = 1 \),- \( A = 1.5 \ \mathrm{m^2} \),- \( \sigma = 5.67 \times 10^{-8} \ \mathrm{W/m^2 \cdot K^4} \),- \( T = 47 + 273 \mathrm{K} \).Thus:\[ (47 + 273)^{4} \approx 127,102,362.96 \]\[ Q_{\text{environment}} \approx 5.67 \times 10^{-8} \times 1.5 \times (127,102,362.96) \approx 1081.5 \ \mathrm{W} \]

Step 5: Calculate Net Heating Rate

The net heat rate increases the skin temperature by adding all sources and subtracting any cooling: \[ Q_{\text{net}} = Q_{\text{metabolic}} + Q_{\text{convection}} + Q_{\text{sun}} + Q_{\text{environment}} \]Substitute calculated and known values:\[ Q_{\text{net}} = 280 + 0.2475 + 2100 + 1081.5 \approx 3461.75 \ \mathrm{W} \]

Step 6: Calculate Perspiration Rate (No Clothing)

Perspiration is the heat loss via evaporation provided heat transfer balances. Using: \[ Q_{\text{perspiration}} = Q_{\text{net}} = m_{\text{evap}} L_{v} \] This formula solves for the mass of evaporated perspiration \( m_{\text{evap}} \):\[ m_{\text{evap}} = \frac{Q_{\text{net}}}{L_{v}} \] With \( L_{v} = 2.42 \times 10^{6} \ \mathrm{J/kg} \), applies:\[ m_{\text{evap}} = \frac{3461.75}{2.42 \times 10^{6}} \approx 0.00143 \ \mathrm{kg/s} \]Convert mass rate to volume (using water density \(1 \mathrm{kg/L} \)) and hourly rate:\[ V_{\text{evap}} \approx 0.00143 \times 3600 \approx 5.15 \ \mathrm{L/h} \]

Step 7: Calculate Perspiration Rate (With Clothing)

Using the same formula but substituting the skin area: \( A_{\text{skin}} = 0.45 \ \mathrm{m^2} \), recalculate heat absorbed from the sun and environment:\[ Q_{\text{sun}} = 1400 \times 0.45 = 630 \ \mathrm{W} \] and \[ Q_{\text{environment}} = 5.67 \times 10^{-8} \times 0.45 \times (127,102,362.96) \approx 324.45 \ \mathrm{W} \]Net heat rate now \[ Q_{\text{net}} = 280 + 0.2475 + 630 + 324.45 \approx 1234.7 \ \mathrm{W} \]Then calculate \[ m_{\text{evap}} = \frac{1234.7}{2.42 \times 10^{6}} \approx 0.00051 \ \mathrm{kg/s} \]Volume rate \[ V_{\text{evap}} \approx 0.00051 \times 3600 \approx 1.84 \ \mathrm{L/h} \]Light-colored clothing significantly decreases the evaporation rate needed, offering protection against heat.

Key Concepts

Metabolic Heat Generation

Metabolic heat generation is a process where the body produces heat through metabolic reactions. When we eat food, our body converts it into energy. During this process, some of the energy is converted into heat, warming our bodies from the inside. This is especially significant in maintaining body temperature in environments where external conditions could lead to heat loss or gain.

In the context of the exercise, the lost soul's body generates 280 W of heat metabolically. This heat accounts for the body's natural energy output at rest or during activity. It's an essential component of the body's internal temperature regulation, ensuring that tissues remain at optimal working temperatures for enzymatic and physiological processes. Apart from maintaining bodily functions, this heat plays a crucial part in balancing other heat transfer processes like convection or radiation.

Here are key points about metabolic heat:

• Originates from cellular activities converting food into energy
• Helps maintain a steady core temperature
• Provides a baseline heat input into the body's heat balance

Convection Heat Transfer

Convection heat transfer occurs when heat is transported through a fluid – usually air or water – by the movement of the fluid itself. This is much like the breeze you feel when a fan is turned on, helping to cool your skin by carrying heat away. In biological systems, convection plays an integral role, especially in regulating skin temperature.

The exercise demonstrates convection where the air temperature influences skin temperature. The rate of heat gain or loss is given by the formula: \[ Q_{\text{convection}} = k^{\prime} A_{\text{skin}} (T_{\text{air}} - T_{\text{skin}}) \]This equation shows that larger differences in air and skin temperatures result in higher convective heat transfer, either warming or cooling the skin. If the surrounding air is hotter than the skin, heat will flow into the body, increasing body temperature. Conversely, if it’s cooler, heat will exit the body through the skin, which can help cool down.

Consider these aspects:

• Dependent on the temperature difference between skin and air
• Regulated by factors like airflow and exposure area
• Exhibits bidirectional heat transfer – could warm or cool the skin
• In the problem, contributes a small amount of heat (0.2475 W) compared to other processes

Radiant Heat Transfer

Radiant heat transfer encompasses the process by which energy is emitted or absorbed in the form of electromagnetic waves, especially infrared radiation. This mechanism does not rely on any medium, meaning heat can be transferred through a vacuum – such as the energy from the sun reaching the Earth.

In our exercise, radiant heat is exceedingly relevant, with two components to consider: direct solar radiation and environmental radiation. The skin absorbs radiation from the sun quantified as 1400 W/m² and from the surrounding at ambient temperature.

The calculation of these components uses the formula: \[ Q_{\text{radiant}} = e \cdot \sigma A (T_{\text{source}} + 273)^4 \]where emissivity \( e \) and Stefan–Boltzmann constant \( \sigma \) play crucial roles in determining total absorbed or emitted radiation. In this scenario, adding the effects together results in a significant amount of energy being absorbed by our lost soul, requiring balancing through other means such as sweating to prevent overheating.

Key points include:

• Involves transfer of energy via infrared waves
• Can occur across vacuum as it doesn't require a medium
• Direct solar irradiance is a major relevant factor on sunny days
• Significant in desert environments, increasing heat absorbed

Evaporative Cooling

Evaporative cooling is an essential process for temperature regulation in living organisms, especially under hot conditions. This process involves the transformation of liquid (sweat) into vapor, which absorbs substantial amounts of heat energy from the body, thereby cooling it down.

In the exercise, perspiration's evaporative cooling effect counters the heat gained from metabolic processes and environmental influences. The rate at which this needs to happen is crucial for maintaining a stable skin temperature during hot conditions. The person's skin requires water to evaporate at a rate that compensates for the net heat gain calculated from all other heat transfer mechanisms.

The formula for evaporative cooling in this context is: \[ Q_{\text{perspiration}} = m_{\text{evap}} \cdot L_v \]where \( m_{\text{evap}} \) is the mass of evaporated perspiration, and \( L_v \) is the latent heat of vaporization of water. In the problem given, the person had to lose around 5.15 L/h through perspiration initially, which could be reduced to 1.84 L/h by wearing light-colored clothing.

Considerations include:

• Relies on water loss through sweat
• Helps in efficiently cooling the body
• Affected by factors like humidity, which impacts evaporation rate
• In desert environments, critical to prevent overheating and dehydration