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Overall, \(80 \%\) of the energy used by the body must be eliminated as excess thermal energy and needs to be dissipated. The mechanisms of elimination are radiation, evaporation of sweat ( \(2430 \mathrm{~kJ} / \mathrm{kg}\) ), evaporation from the lungs ( \(38 \mathrm{~kJ} / \mathrm{h}\) ), conduction, and convection. A person working out in a gym has a metabolic rate of \(2500 \mathrm{~kJ} / \mathrm{h}\). His body temperature is \(37^{\circ} \mathrm{C}\), and the outside temperature \(24^{\circ} \mathrm{C}\). Assume the skin has an area of \(2.0 \mathrm{~m}^{2}\) and emissivity of \(0.97\). (a) At what rate is his excess thermal energy dissipated by radiation? (b) If he eliminates \(0.40 \mathrm{~kg}\) of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (c) At what rate is energy eliminated by evaporation from the lungs? (d) At what rate must the remaining excess energy be eliminated through conduction and convection?

Short Answer

Expert verified
The steps above detail how to calculate the rates of energy dissipation by: radiation, sweat evaporation, lung evaporation, and the combination of conduction and convection. The key steps involve converting all measurements to a common unit and methodically following the laws of thermodynamics.

Step by step solution

01

Calculate the rate of dissipation by radiation.

The formula to calculate the power radiated is given by the Stefan-Boltzmann law: \( P = \varepsilon \sigma A T^{4} \) where \( \varepsilon \) is the emissivity, The Stefan-Boltzmann constant is \( \sigma = 5.67 × 10^{-8} W/m^{2} K^{4}\), \( A \) is the surface area and \( T \) is the absolute temperature. We first convert the body temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature. Plugging in the known values we get \( P_{rad} = 0.97*5.67 × 10^{-8} W/m^{2} K^{4}*2 m^{2}*(310.15 K)^{4}\)
02

Calculate the rate of dissipation through evaporation of sweat.

The energy removed by sweat is given by the equation \( Q = mL \), where \( m \) is the mass of the sweat and \( L \) is the latent heat of evaporation. We convert this energy into a rate by dividing the result by the amount of time (1 hour or 3600 seconds). So, \( P_{sweat}= \frac{0.4 kg*2430 kJ/kg}{3600s} \).
03

Calculate the rate of dissipation through evaporation from the lungs.

It is stated in the problem that respiration removes \( 38 kJ/h \), hence no calculations are necessary. However, the rate in terms of seconds is needed, so the rate of dissipation through the lungs is \( P_{lungs}= \frac{38 kJ}{3600s} \).
04

Determine the rate of elimination through conduction and convection.

The total excess energy produced by the body is the metabolic rate, which is 80% of 2500 kJ/h. Subtract the energy losses calculated in the previous steps from the total excess energy. Then convert this remaining energy into a rate by dividing by the number of seconds in an hour. So, \( P_{con} = \frac{0.8*2500 kJ - (P_{rad} + P_{sweat}+ P_{lungs})}{3600s} \).

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Key Concepts

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

Stefan-Boltzmann law
The Stefan-Boltzmann law is a key principle in understanding how objects radiate thermal energy. This law states that the power radiated per unit area of an object is proportional to the fourth power of its temperature in Kelvin. In mathematical terms, it is expressed as:
  • \( P = \varepsilon \sigma A T^{4} \)
Where \( P \) is the power radiated, \( \varepsilon \) is the emissivity, \( \sigma \) the Stefan-Boltzmann constant \((5.67 × 10^{-8} \text{ W/m}^{2} \text{K}^{4})\), \( A \) is the surface area, and \( T \) is the absolute temperature.
In our example, the human body is radiating heat due to the difference in temperature between body surface (37°C) and the surrounding environment (24°C). By converting these temperatures into Kelvin and using the given emissivity and surface area, the rate at which thermal energy is dissipated by radiation can be calculated. This calculation is crucial for understanding how much thermal energy needs to be removed from the body to maintain a stable internal environment.
Evaporation of sweat
Sweating is an essential mechanism the body uses to regulate temperature. When you sweat, the water on the skin surface absorbs heat from the body and evaporates, thus cooling you down. The energy required for this evaporation is known as latent heat, and for sweat, it takes about 2430 kJ to evaporate 1 kg of water.
To find out how much thermal energy is removed from the body through the evaporation of sweat, we use the equation:
  • \( Q = mL \)
Where \( m \) is the mass of sweat and \( L \) is the latent heat of evaporation per unit mass. In the gym example, if a person sweats out 0.40 kg of water in an hour, this energy can be calculated to yield the rate of thermal energy dissipation. This process effectively helps in maintaining the body’s thermal balance.
Metabolic rate
The metabolic rate is the rate at which your body consumes energy to maintain vital functions and perform activities. In the context of thermal energy dissipation, the metabolic rate represents the total energy produced by the body. In the given scenario, the person has a metabolic rate of 2500 kJ/h.
Since 80% of this energy is released as heat, it needs to be dissipated to maintain a consistent internal temperature. This heat is eliminated through various mechanisms such as radiation, evaporation, conduction, and convection. Understanding the metabolic rate is essential for calculating how much energy needs to be managed by these processes. It provides a basis to determine how the body efficiently removes excess heat.
Emissivity
Emissivity is a measure of how effectively a surface emits thermal radiation compared to an ideal black body. It ranges from 0 to 1, where 1 represents a perfect emitter. The concept of emissivity is important in calculating the radiative heat loss from the human body using the Stefan-Boltzmann law.
In our example, the person's skin has an emissivity of 0.97, meaning it is very close to being a perfect emitter. This high emissivity factor signifies that the skin is efficient at radiating away excess heat. By knowing the emissivity, one can determine how much heat is lost through radiation, which is a critical component in managing the body's energy balance during physical activities such as working out in a gym.

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Most popular questions from this chapter

A sprinter of mass \(m\) accelerates uniformly from rest to velocity \(v\) in \(t\) seconds. (a) Write a symbolic expression for the instantaneous mechanical power \(P\) required by the sprinter in terms of force \(F\) and velocity v. (b) Use Newton's second law and a kinematics equation for the velocity at any time to obtain an expression for the instantaneous power in terms of \(m, a\), and tonly (c) If a \(75.0-\mathrm{kg}\) sprinter reaches a speed of and \(t\) only. (c) If a \(75.0-\mathrm{kg}\) sprinter reaches a speed of \(11.0 \mathrm{~m} / \mathrm{s}\) in \(5.00 \mathrm{~s}\), calculate the sprinter's acceleration, \(11.0 \mathrm{~m} / \mathrm{s}\) in \(5.00 \mathrm{~s}\), calculate the sprinter's acceleration, assuming it to be constant. (d) Calculate the \(75.0-\mathrm{kg}\) sprinter's instantaneous mechanical power as a function of time \(t\) and (e) give the maximum rate at which he burns Calories during the sprint, assuming \(25 \%\) efficiency of conversion form food energy to mechanical energy.

A thermopane window consists of two glass panes, each \(0.50 \mathrm{~cm}\) thick, with a \(1.0-\mathrm{cm}\)-thick sealed layer of air in between. (a) If the inside surface temperature is \(23^{\circ} \mathrm{C}\) and the outside surface temperature is \(0.0^{\circ} \mathrm{C}\), determine the rate of energy transfer through \(1.0 \mathrm{~m}^{2}\) of the window. (b) Compare your answer to (a) with the rate of energy transfer through \(1.0 \mathrm{~m}^{2}\) of a single \(1.0-\mathrm{cm}\) thick pane of glass. Disregard surface air layers.

The average thermal conductivity of the walls (including windows) and roof of a house in Figure P11.42 is \(4.8 \times 10^{-4} \mathrm{~kW} / \mathrm{m}^{\circ}{ }^{\circ} \mathrm{C}\), and their average thickness is \(21.0 \mathrm{~cm}\). The house is heated with natural gas, with a heat of combustion (energy released per cubic meter of gas burned) of \(9300 \mathrm{kcal} / \mathrm{m}^{3}\). How many cubic meters of gas must be burned each day to maintain an inside temperature of \(25.0^{\circ} \mathrm{C}\) if the outside temperature is \(0.0^{\circ} \mathrm{C}\) ? Disregard surface air layers, radiation, and energy loss by heat through the ground.

When a driver brakes an automobile, the friction between the brake drums and the brake shoes converts the car's kinetic energy to thermal energy. If a \(1500-\mathrm{kg}\) automobile traveling at \(30 \mathrm{~m} / \mathrm{s}\) comes to a halt, how much does the temperature rise in each of the four \(8.0-\mathrm{kg}\) iron brake drums? (The specific heat of iron is \(448 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\).)

The apparatus shown in Figure P11.10 was used by Joule to measure the mechanical equivalent of heat. Work is done on the water by a rotating paddle wheel, which is driven by two blocks falling at a constant speed. The temperature of the stirred water increases due to the friction between the water and the paddles. If the energy lost in the bearings and through the walls is neglected, then the loss in potential energy associated with the blocks equals the work done by the paddle wheel on the water. If each block has a mass of \(1.50 \mathrm{~kg}\) and the insulated tank is filled with \(200 \mathrm{~g}\) of water, what is the increase in temperature of the water after the blocks fall through a distance of \(3.00 \mathrm{~m}\) ?

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