91Ó°ÊÓ

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass 68 kg and surface area 1.85 m\(^2\) produces energy at a rate of up to 1300 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\(^\circ\)C instead of the usual 30\(^\circ\)C. (Ignore 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\(^\circ\)C (104\(^\circ\)F)? (Remember: 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 his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (e) How many 750-mL bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 kg.

Step by step solution

01

Calculate Heat Produced by Jogging

Given that the runner produces energy at a rate of 1300 W during jogging, 80% of this energy is converted into heat. Therefore, we calculate the heat produced per second as follows:\[ Q_{jog} = 0.80 \times 1300 \, \text{W} = 1040 \, \text{J/s} \]

02

Calculate Net Heat Gain from Radiation

The amount of heat radiated by the jogger is calculated using the Stefan-Boltzmann law, adjusted for environmental radiation:Radiated heat by body:\[ Q_{out} = \sigma \epsilon A (T_s^4) \]Absorbed heat from environment:\[ Q_{in} = \sigma \epsilon A (T_a^4) \]Where:- \( \sigma = 5.67 \times 10^{-8} \text{W/m}^2\text{K}^4 \) (Stefan-Boltzmann constant)- \( \epsilon = 1 \) (assumed emissivity)- \( A = 1.85 \text{ m}^2 \) (surface area)- \( T_s = 306 \text{ K} \) (body surface temperature)- \( T_a = 313 \text{ K} \) (air temperature)Net heat gained:\[ Q_{net} = \sigma A (T_s^4 - T_a^4) = 5.67 \times 10^{-8} \times 1.85 \times (306^4 - 313^4) \]Calculate and simplify to get approximately:\[ Q_{net} \approx -138 \, \text{J/s} \]

03

Calculate Total Excess Heat Per Second

The total excess heat that the runner needs to dissipate is the sum of the metabolic heat production and the net heat gain from radiation:\[ Q_{excess} = Q_{jog} + Q_{net} = 1040 \, \text{J/s} + (-138 \, \text{J/s}) = 902 \, \text{J/s} \]

04

Calculate Water Evaporated Every Minute

The amount of water evaporated to dissipate the excess heat can be calculated using the heat of vaporization:\[ \text{Heat required to evaporate water in 1 second:} \, Q_{excess} = m L_v \]\[ m = \frac{Q_{excess}}{L_v} = \frac{902 \, \text{J/s}}{2.42 \times 10^6 \, \text{J/kg}} \approx 0.0003734 \, \text{kg/s} \]Convert this to kg per minute by multiplying by 60 seconds:\[ m_{minute} = 0.0003734 \, \text{kg/s} \times 60 \, \text{s} = 0.0224 \, \text{kg/min} \]

05

Calculate Number of Water Bottles Required

To find out how many 750 mL bottles are needed after jogging for half an hour, calculate the total water loss:\[ \text{Total water loss over 30 minutes:} = 0.0224 \, \text{kg/min} \times 30 \, \text{min} = 0.672 \, \text{kg} \]Since 750 mL of water has a mass of 0.75 kg:\[ \text{Number of bottles required:} = \frac{0.672}{0.75} \approx 0.896 \]Thus, approximately 1 bottle is required.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Heat Transfer

Heat transfer plays a crucial role in thermal physics, especially when it involves the human body's ability to stay cool during activities like jogging. There are three primary modes of heat transfer: conduction, convection, and radiation. However, in this case, we focus on radiation since conduction was ignored and convection is not directly accounted for.
When a person jogs, their body generates heat. This heat needs to be dissipated to maintain a stable body temperature. The process of sweating and evaporating water from the skin is a key method for cooling down through heat transfer. This process involves the movement of heat energy from the body into the surrounding air, mainly through radiation and convection, when moving air carries away the heat.
Understanding how the body balances heat production and loss is essential for activities in hot climates. Proper hydration helps in efficient heat transfer through sweating, mitigating the heat stress experienced during vigorous exercise.

Black Body Radiation

Black body radiation is a concept in which a perfect emitter and absorber of radiation is considered. In thermal physics, black bodies are objects that absorb all incoming radiation without reflecting any. Humans, although not perfect black bodies, do radiate heat as black bodies do.
Using the Stefan-Boltzmann Law, we can understand how the body releases heat through radiation. The law is expressed as:

• \[ Q = \sigma \epsilon A T^4 \]

Where:

• \(Q\) is the radiated heat energy.
• \(\sigma\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \, ext{W/m}^2\text{K}^4.\)
• \(\epsilon\) is the emissivity, which describes how efficient a surface is at emitting thermal radiation (in this scenario, assumed to be 1 for maximum efficiency).
• \(A\) is the surface area of the body, and \(T\) is the temperature.

The jogger radiates heat at their body temperature but also absorbs some from the surroundings, which has implications for their thermal comfort. Understanding black body radiation helps in quantifying the heat balance for alleviating stress from intense exercises.

Heat of Vaporization

The heat of vaporization refers to the amount of energy required to turn a given amount of liquid into vapor without a temperature change. For water, this is particularly important in the process of sweating, one of the body's natural cooling methods.
In thermal terms, when the body sweats, it releases heat by evaporating water from the skin's surface, which consumes energy measured as the heat of vaporization. For water at body temperature, this value is approximately \(2.42 \times 10^6 \, ext{J/kg}\).
This concept is vital in calculating how much sweat is needed to remove excess heat to avoid overheating during physical activities. The energy required to evaporate sweat makes it possible for the body to maintain a stable internal temperature even when exercising in hot conditions.
Therefore, understanding the heat of vaporization is key to managing hydration levels and ensuring that the body can effectively utilize sweating for temperature regulation.

Emissivity

Emissivity is the measure of an object's ability to emit thermal radiation. It is a dimensionless quantity that ranges from 0 to 1, with 1 being a perfect emitter. In our example, the runner is assumed to have an emissivity of 1, meaning their body efficiently radiates heat like a black body.
Emissivity affects how much heat the body can lose through radiation. A higher emissivity means better heat dissipation, crucial for maintaining body temperature during high temperatures or extensive physical activity.
In practical scenarios, while human skin is not a perfect black body, it has a high emissivity value typically above 0.9. This allows the body to effectively release heat, significantly aiding in temperature regulation.
Understanding emissivity helps when designing clothing and environments that aid or block heat dissipation. For joggers, wearing loose, breathable clothing can enhance emissivity effects, allowing better adaptation to thermal conditions.