/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Even at rest, the human body gen... [FREE SOLUTION] | 91Ó°ÊÓ

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Even at rest, the human body generates heat. The heat arises because of the body's metabolism - that is, the chemical reactions that are always occurring in the body to generate energy. In rooms designed for use by large groups, adequate ventilation or air conditioning must be provided to remove this heat. Consider a classroom containing 200 students. Assume that the metabolic rate of generating heat is \(130 \mathrm{~W}\) for each student and that the heat accumulates during a fifty-minute lecture. In addition, assume that the air has a molar specific heat of \(C_{V}=\frac{5}{2} R\) and that the room (volume \(=1200 \mathrm{~m}^{3}\), initial pressure \(=1.01 \times 10^{5} \mathrm{~Pa}\), and intial temperature \(=21^{\circ} \mathrm{C}\) ) is sealed shut. If all the heat generated by the students were absorbed by the air, by how much would the air temperature rise during a lecture?

Short Answer

Expert verified
The air temperature rises by approximately 30.2 K during the lecture.

Step by step solution

01

Determine Total Heat Generated

First, calculate the total heat generated by all students in the classroom. Each student generates heat at a rate of 130 W. For 200 students, this becomes:\[Q_{total} = 200 \times 130 \text{ W} = 26000 \text{ W}\]The class lasts for 50 minutes, or \(50 \times 60 = 3000\) seconds. Thus, the total heat generated during this time is:\[Q = 26000 \text{ W} \times 3000 \text{ s} = 7.8 \times 10^7 \text{ J}\]
02

Determine Moles of Air in the Room

The room volume is \(1200\ \text{m}^3\). Using the ideal gas law \(PV = nRT\), calculate the number of moles \(n\). We need to express the temperature \(T\) in Kelvin, so \(T = 21 + 273.15 = 294.15\ \text{K}\).Rearranging the ideal gas law gives:\[n = \frac{PV}{RT}\]Substituting the given values:\[n = \frac{1.01 \times 10^5 \times 1200}{8.31 \times 294.15} \approx 49643.9\text{ moles}\]
03

Calculate Temperature Increase

The total heat absorbed by the air is given by the formula \(Q = nC_V\Delta T\), where \(C_V = \frac{5}{2}R\). Solving for the temperature increase \(\Delta T\):\[Q = n\left(\frac{5}{2}R\right)\Delta T\Rightarrow \Delta T = \frac{Q}{n\left(\frac{5}{2}R\right)}\]Substitute the known quantities:\[\Delta T = \frac{7.8 \times 10^7}{49643.9 \times \left(\frac{5}{2} \times 8.31\right)}\approx 30.2\ \text{K}\]Thus, the air temperature would rise by approximately 30.2 K.

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Key Concepts

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

Metabolic Heat Generation
Metabolic heat generation is a fascinating process where heat is produced in the human body through metabolic reactions. These are the chemical reactions that convert the food we consume into energy. This energy is not just used for mechanical work but also generates heat as a byproduct.
For students sitting in a classroom, each generates about 130 watts (W) of metabolic heat simply by existing!
If you consider a large group, like 200 students, the total heat generated is substantial. Calculating it in our example, we find that the total heat generated is 26,000 W per minute. Over the span of a typical 50-minute lecture, this means nearly 7.8 million joules (J) of heat is produced and accumulated if not properly ventilated or removed.
Understanding metabolic heat generation helps architects and engineers design efficient heating and cooling systems in buildings, especially areas meant for large groups, like classrooms and lecture halls.
Ideal Gas Law
The ideal gas law is a fundamental equation that connects various properties of gases: pressure (P), volume (V), temperature (T), and the number of moles (n). By the formula: \[ PV = nRT \] where R is the ideal gas constant.
This law helps us understand how these properties interact.
  • Pressure: the force the gas exerts on the walls of its container.
  • Volume: the space the gas occupies.
  • Temperature: a measure of the average kinetic energy of the gas molecules.
  • Moles: the amount of gas present.
In the classroom scenario, where air is treated as an ideal gas, we used this law to determine how many moles of air filled the sealed room.
By manipulating the equation for our known quantities (initial pressure, room volume, and temperature), we calculated approximately 49,643.9 moles of air are present. This step is crucial for determining how much heat the air can absorb and, ultimately, how much the temperature rises due to students' metabolic heat.
Specific Heat Capacity
Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). However, when dealing with gases, we often consider molar specific heat, which relates to temperature change per mole rather than per mass unit.
In our scenario, the selected molar specific heat is for constant volume, indicated as \( C_V = \frac{5}{2}R \), where \( R \) is the universal gas constant.
Understanding specific heat capacity is essential since it dictates how a substance responds to the input of energy.
In the context of the classroom problem, it informs us how much heat the air can absorb without significant temperature change.
Using the calculated moles of air and the total heat produced, we solved for the temperature increase \( \Delta T \) using the relationship: \[ Q = nC_V\Delta T \] This relationship allowed us to determine that the air temperature would rise by approximately 30.2 Kelvin during a lecture due to the students' metabolic heat.
Heat Transfer
Heat transfer is all about the movement of thermal energy from one place or material to another. There are three primary modes: conduction, convection, and radiation. However, the classroom scenario primarily involves convection.
In a closed space like a lecture hall, the heat generated by students' metabolism may cause the air temperature to rise if the heat isn't effectively removed.
This setting requires proper ventilation to transport heat away and maintain a comfortable environment.
Let's break down some key points:
  • Conduction: Transfer of heat through direct contact.
  • Convection: Transfer of heat by the movement of fluids or gases.
  • Radiation: Transfer of heat in the form of electromagnetic waves.
Here, knowing the amount of heat generated, and the resultant temperature rise helps in planning adequate ventilation systems.
By maintaining efficient heat transfer, the environment can remain stable and comfortable, ensuring both safety and a conducive atmosphere for students. In our problem, without proper heat transfer mechanisms, the temperature was found to rise significantly within the sealed room over the course of the lecture.

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Most popular questions from this chapter

A system gains a certain amount of energy in the form of heat at constant pressure, and the internal energy of the system increases by an even greater amount. (a) Is any work done and, if so, is it done on or by the system? (b) If there is work, is it positive or negative, according to our convention? (c) Does the volume of the system increase, decrease, or remain the same? Give your reasoning A system gains \(2780 \mathrm{~J}\) of heat at a constant pressure of \(1.26 \times 10^{5} \mathrm{~Pa}\) and its internal energy increases by \(3990 \mathrm{~J}\). What is the change in volume of the system, and is it an increase or a decrease? Verify that your answer is consistent with your answers to the Concept Questions.

The inside of a Carnot refrigerator is maintained at a temperature of \(277 \mathrm{~K},\) while the temperature in the kitchen is \(299 \mathrm{~K}\). Using \(2500 \mathrm{~J}\) of work, how much heat can this refrigerator remove from its inside compartment?

The drawing (not to scale) shows the way in which the pressure and volume change for an ideal gas that is used as the working substance in a Carnot engine. The gas begins at point a (pressure \(=P_{\mathrm{a}}\), volume \(\left.=V_{\mathrm{a}}\right)\) and expands isothermally at temperature \(T_{\mathrm{H}}\) until point \(\mathrm{b}\) (pressure \(=P_{\mathrm{b}}\), volume \(=V_{\mathrm{b}}\) ) is reached. During this expansion, the input heat of magnitude \(\left|Q_{\mathrm{H}}\right|\) enters the gas from the hot reservoir of the engine. Then, from point \(\mathrm{b}\) to point \(\mathrm{c}\) (pressure \(=P_{\mathrm{c}}\), volume \(\left.=V_{\mathrm{c}}\right)\), the gas expands adiabatically. Next, the gas is compressed isothermally at temperature \(T_{\mathrm{C}}\) from point \(\mathrm{c}\) to point \(\mathrm{d}\) (pressure \(=P_{\mathrm{d}}\), volume \(=V_{\mathrm{d}}\) ). During this compression, heat of magnitude \(\left|Q_{\mathrm{C}}\right|\) is rejected to the cold reservoir of the engine. Finally, the gas is compressed adiabatically from point \(d\) to point a, where the gas is back in its initial state. The overall process a to \(b\) to \(\mathrm{c}\) to \(\mathrm{d}\) is called a Carnot cycle. Prove for this cycle that \(|Q \mathrm{c}| /|Q \mathrm{H}|=T_{\mathrm{C}} / T_{\mathrm{H}}\)

A mountain climber, starting from rest, does work in climbing upward. At the if top, she is again at rest. In the process, her body generates \(4.1 \times 10^{6} \mathrm{~J}\) of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by Equation 15.11 as \(e=|W| /\left|Q_{\mathrm{H}}\right|,\) where \(|W|\) is the magnitude of the work and \(\left|Q_{\mathrm{H}}\right|\) is the magnitude the input heat. (a) Is the \(4.1 \times 10^{6} \mathrm{~J}\) of energy equal to \(|W|\) or to \(\left|Q_{\mathrm{H}}\right| ?\) (b) How is the work done in climbing upward related to the vertical height of the climb? Explain. Problem The vertical height of the climb is \(730 \mathrm{~m}\). The climber has a mass of \(52 \mathrm{~kg}\). Find her efficiency as a heat engine.

Three moles of an ideal monatomic gas are at a temperature of \(345 \mathrm{~K}\). Then, \(2438 \mathrm{~J}\) of heat is added to the gas, and \(962 \mathrm{~J}\) of work is done on it. What is the final temperature of the gas?

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