Problem 117 - University Physics with Modern Physics

Question

Jogging in the Heat of the Day. You have probably seen people jogging in extremely hot weather and wondered Why? As we shall see, there are good reasons not to do this! When jogging strenuously, an average runner of mass 68 \(\mathrm{kg}\) and surface area 1.85 \(\mathrm{m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{W}, 80 \%\) of which is converted to heat. The jogger radiates heat, but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C}\) . (We shall neglect conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0 \(^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right)\) ? (Remember that he radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\). must get rid of per second? (d) How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}.\)

Step-by-Step Solution

Answer

(a) 1040 W, (b) -62.66 W, (c) 977.34 W, (d) 0.024 kg/min.

Step 1: Calculate the Heat Produced by Jogging

To find the heat produced by jogging, we need to consider that 80% of the energy produced during jogging is converted to heat. The rate at which energy is produced is given as 1300 W (watts). Thus, the amount of heat produced per second is \[Q_{ ext{produced}} = 0.80 \times 1300 \, \text{W} = 1040 \, \text{W}.\]

Step 2: Determine the Net Heat Gain from Radiation

The heat radiated by the jogger and absorbed from the environment is considered here. Using the Stefan-Boltzmann Law, the net radiative heat gain can be calculated: \[P_{ ext{net}} = A \varepsilon \sigma \left( T_{ ext{env}}^4 - T_{ ext{jogger}}^4 \right),\]where: - \( A = 1.85 \, \text{m}^2 \) is the surface area,- \( \varepsilon = 1 \) (assumed for simplicity, maximal emissivity),- \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \) is the Stefan-Boltzmann constant,- \( T_{ ext{env}} = 313.15 \, \text{K} \) (40°C),- \( T_{ ext{jogger}} = 306.15 \, \text{K} \) (33°C).Substituting the values,\[P_{ ext{net}} = 1.85 \times 1 \times 5.67 \times 10^{-8} \times (313.15^4 - 306.15^4) \= -62.66 \, \text{W}.\] The negative sign indicates net heat is gained from the environment.

Step 3: Calculate Total Excess Heat

The total excess heat that needs to be removed per second is the sum of the heat produced from jogging and the net gain from radiation:\[Q_{ ext{excess}} = Q_{ ext{produced}} + P_{ ext{net}} = 1040 + (-62.66) \approx 977.34 \, \text{W}.\]

Step 4: Find the Amount of Water Evaporated Per Minute

To find how much water must be evaporated to remove the excess heat, we use the heat of vaporization:\[Q_{ ext{minute}} = Q_{ ext{excess}} \times 60 = 977.34 \times 60 = 58640.4 \, \text{J/min}. \]The amount of water evaporated every minute is given by:\[m = \frac{Q_{ ext{minute}}}{L_v} = \frac{58640.4}{2.42 \times 10^6} \approx 0.024 \, \text{kg/min}. \]

Key Concepts

Heat transfer

Heat transfer is a fundamental concept in thermodynamics, which deals with the movement of thermal energy from one object to another. This transfer can happen through three primary mechanisms: conduction, convection, and radiation. In our scenario, conduction is ignored, as it's not the main player in the jogger's heat balance.

Instead, radiation and convection are crucial. Here, the jogger's skin and the surrounding environment are continuously exchanging heat. Heat moves from areas of higher temperature to lower temperature, and since the jogger is warmer than the surrounding air, heat radiates away from his body. This process can be quantified with specific laws that govern heat transfer.

Keeping track of how much energy is transferred or gained helps to determine the amount of heat that needs to be managed by the body through other processes, such as sweating.

Sweating as a cooling mechanism

Sweating is the body's natural way to regulate temperature, particularly under strenuous activities like jogging. When the body heats up, it releases sweat—a mixture mostly made up of water—onto the skin's surface. The key cooling effect occurs when the sweat evaporates.

Evaporation is an endothermic process, meaning it requires heat to proceed. The body supplies this heat, which is taken away as water transitions from liquid to gas. Thus, sweating facilitates the removal of excess body heat, helping to cool the body down.

This process isn't only active during exercise but is incredibly vital when external temperatures make it difficult for the body to lose heat through radiation or convection. Thus, when you notice someone sweating profusely after a jog, it's their body's way of ensuring their internal temperature doesn't rise to dangerous levels.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a principle in physics that relates the radiated heat energy of a body to its temperature. Specifically, it says that the power emitted per unit area of a black body is proportional to the fourth power of its absolute temperature.

The equation is expressed as: \[ P = A \, \varepsilon \, \sigma \left(T^4 \right) \]where:

• \( P \) is the power radiated per unit area.
• \( A \) is the surface area of the body.
• \( \varepsilon \) is the emissivity of the body, which is a measure of how efficiently a surface emits thermal radiation compared to a perfect black body.
• \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \).
• \( T \) is the absolute temperature in Kelvin.

In the case of the jogger, this law helps quantify the net heat exchanged between him and the environment, factoring in both the heat he emits and the heat absorbed from the surroundings.

Jogging heat production

While jogging, a significant amount of energy is produced within the body, and much of it ends up as heat. Heat production is an inevitable byproduct of the muscular activity involved in jogging.

For the jogger in question, the energy production rate given is 1300 W. However, only 80% of this energy is actually transformed into heat, while the rest goes to maintaining movement and other physiological functions. Thus, the heat produced can be calculated and is key to understanding the temperature regulation needed.

This heat must be effectively removed from the body, either through radiation, convection, or evaporation, to maintain a safe body temperature.

Heat of vaporization

The heat of vaporization is the energy required to convert a liquid into a gas at constant temperature and pressure. For the jogger, this specifically refers to the energy needed for the sweat to evaporate from the skin's surface.

The heat of vaporization for water at body temperature is approximately \( 2.42 \times 10^6 \mathrm{J} / \mathrm{kg} \). This means it takes 2.42 million Joules to evaporate 1 kg of water.

This concept is crucial because it provides a quantitative measure of how effective sweating is in cooling the body. It allows us to calculate how much water needs to be evaporated to rid the body of the excess heat generated from jogging, ensuring that the jogger maintains a stable internal environment.