Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 27

Imagine that 501 people are present in a movie theater of volume $8.00 \times 10^{3} \mathrm{m}^{3}$ that is sealed shut so no air can escape. Each person gives off heat at an average rate of \(110 \mathrm{W} .\) By how much will the temperature of the air have increased during a 2.0 -h movie? The initial pressure is \(1.01 \times 10^{5} \mathrm{Pa}\) and the initial temperature is \(20.0^{\circ} \mathrm{C} .\) Assume that all the heat output of the people goes into heating the air (a diatomic gas).

Short Answer

Expert verified
Based on the given information and calculations, the increase in temperature of the air inside the sealed movie theater after the 2-hour duration of the movie is approximately 9.52 K.

Step by step solution

01

Find the total heat output of the people

First, we need to find the total heat output from all the people after the 2-hour duration of a movie. Multiply the average heat output per person by the number of people and the duration of the movie, which we must convert to seconds: Total heat output = (Number of people) × (average heat output per person) × (time in seconds) Total heat output = 501 × 110 W × (2.0 hours × 3600 s/hour) = 3.32 × 10^7 J
02

Use the ideal gas law to find the initial number of moles in the theater

We need to find the number of moles of air inside the theater initially so that we can calculate the change in temperature due to the heat added. To do this, we will use the ideal gas law, as follows: PV = nRT Here, P = Initial pressure = 1.01 × 10^5 Pa V = Volume of theater = 8.00 × 10^3 m³ R = Ideal gas constant = 8.314 J/(mol K) T = Initial temperature, which we must convert to Kelvin: 20.0°C + 273.15 = 293.15 K Now, solve for the number of moles, n: n = PV/(RT) = (1.01 × 10⁵ Pa)(8.00 × 10³ m³) / (8.314 J/mol K)(293.15 K) = 3.50 × 10⁴ mol
03

Calculate the increase in temperature using the heat capacity of diatomic gases

Now that we know the number of moles and the total heat output, we can use the heat capacity of diatomic gases to calculate the change in temperature. The molar heat capacity at constant volume, C_v, for diatomic gases (air can be considered a diatomic gas) is: C_v = (5/2)R, where R = 8.314 J/(mol K) The total heat added to the system can be expressed as: ΔQ = n × C_v × ΔT Rearrange and solve for the change in temperature, ΔT: ΔT = ΔQ / (n × C_v) = (3.32 × 10⁷ J) / (3.50 × 10⁴ mol × (5/2)(8.314 J/mol K)) = 9.52 K Since the question asks for the increase in temperature, our final answer is: Temperature increase = 9.52 K

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the heat capacity of a system consisting of (a) a \(0.450-\mathrm{kg}\) brass cup filled with \(0.050 \mathrm{kg}\) of water? (b) \(7.5 \mathrm{kg}\) of water in a 0.75 -kg aluminum bucket?
A mass of \(1.00 \mathrm{kg}\) of water at temperature \(T\) is poured from a height of \(0.100 \mathrm{km}\) into a vessel containing water of the same temperature \(T,\) and a temperature change of \(0.100^{\circ} \mathrm{C}\) is measured. What mass of water was in the vessel? Ignore heat flow into the vessel, the thermometer, etc.
The thermal conductivity of the fur (including the skin) of a male Husky dog is \(0.026 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) The dog's heat output is measured to be \(51 \mathrm{W}\), its internal temperature is $38^{\circ} \mathrm{C},\( its surface area is \)1.31 \mathrm{m}^{2},$ and the thickness of the fur is \(5.0 \mathrm{cm} .\) How cold can the outside temperature be before the dog must increase its heat output?
Bare, dark-colored basalt has a thermal conductivity of 3.1 $\mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ whereas light-colored sandstone's thermal conductivity is only \(2.4 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) Even though the same amount of radiation is incident on both and their surface temperatures are the same, the temperature gradient within the two materials will differ. For the same patch of area, what is the ratio of the depth in basalt as compared with the depth in sandstone that gives the same temperature difference?
Many species cool themselves by sweating, because as the sweat evaporates, heat is given up to the surroundings. A human exercising strenuously has an evaporative heat loss rate of about \(650 \mathrm{W}\). If a person exercises strenuously for 30.0 min, how much water must he drink to replenish his fluid loss? The heat of vaporization of water is \(2430 \mathrm{J} / \mathrm{g}\) at normal skin temperature.
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Physics of Motion

Read Explanation

Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Fundamentals of Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free