Problem 74 - College Physics

Question

(a) A typical student listening attentively to a physics lecture has a heat output of \(100 \mathrm{~W}\). How much heat energy does a class of 90 physics students release into a lecture hall over the course of a 50 min lecture? (b) Assume that all the heat energy in part (a) is transferred to the \(3200 \mathrm{~m}^{3}\) of air in the room. The air has a specific heat of \(1020 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K})\) and a density of \(1.20 \mathrm{~kg} / \mathrm{m}^{3} .\) If none of the heat escapes and the air- conditioning system is off, how much will the temperature of the air in the room rise during the 50 min lecture? (c) If the class is taking an exam, the heat output per student rises to \(280 \mathrm{~W}\). What is the temperature rise during \(50 \mathrm{~min}\) in this case?

Step-by-Step Solution

Answer

The temperature rise for listening students is 6.88 K, and for exam-taking students, it is 19.27 K.

Step 1: Calculate Total Heat Output for Listening Students

First, calculate the total heat energy released by 90 students over a 50-minute lecture, given that each has a heat output of 100 W. Calculate the total energy: \[ Q = P \times t \times N \]where \( P = 100 \, \mathrm{W} \), \( t = 50 \times 60 \, \mathrm{s} \) (since 1 minute = 60 seconds), and \( N = 90 \) (number of students). Plug in the values:\[ Q = 100 \, \mathrm{W} \times 3000 \, \mathrm{s} \times 90 \]\[ Q = 27,000,000 \, \mathrm{J} \]

Step 2: Find Temperature Increase from Heat Energy for Listening Students

Use the formula for heat energy transfer to find the temperature increase of the air in the room:\[ \Delta T = \frac{Q}{m \cdot c} \]where \( Q = 27,000,000 \, \mathrm{J} \), \( c = 1020 \, \mathrm{J}/(\mathrm{kg \cdot K}) \), and \( m = \rho \times V \). The mass of air \( m \) is calculated using the density \( \rho = 1.20 \, \mathrm{kg/m}^3 \) and the volume \( V = 3200 \, \mathrm{m}^3 \):\[ m = 1.20 \, \mathrm{kg/m}^3 \times 3200 \, \mathrm{m}^3 = 3840 \, \mathrm{kg} \]Substitute these values into the equation:\[ \Delta T = \frac{27,000,000 \, \mathrm{J}}{3840 \, \mathrm{kg} \times 1020 \, \mathrm{J/(kg \cdot K)}} \]\[ \Delta T = 6.88 \, \mathrm{K} \]

Step 3: Calculate Total Heat Output for Exam Taking Students

When students take an exam, each has a heat output of 280 W. Calculate the total heat energy released by these 90 students over a 50-minute lecture:\[ Q = P \times t \times N \]where \( P = 280 \, \mathrm{W} \), \( t = 50 \times 60 \, \mathrm{s} \), and \( N = 90 \):\[ Q = 280 \, \mathrm{W} \times 3000 \, \mathrm{s} \times 90 \]\[ Q = 75,600,000 \, \mathrm{J} \]

Step 4: Find Temperature Increase from Heat Energy for Exam Taking Students

Using the previously calculated mass and specific heat, find the temperature increase of the air due to exam-taking students:\[ \Delta T = \frac{Q}{m \cdot c} \]where \( Q = 75,600,000 \, \mathrm{J} \):\[ \Delta T = \frac{75,600,000 \, \mathrm{J}}{3840 \, \mathrm{kg} \times 1020 \, \mathrm{J/(kg \cdot K)}} \]\[ \Delta T = 19.27 \, \mathrm{K} \]

Key Concepts

Heat Transfer

Heat transfer is the movement of thermal energy from one object or material to another. In thermodynamics, understanding how heat moves is crucial.
In the context of our exercise, heat transfer occurs when students generate heat, which is absorbed by the air in the lecture hall. This process includes:

• Conduction: Heat moves through direct contact. Not as relevant here since air is less conductive.
• Convection: This is important as warm air rises and circulates. It helps distribute the heat energy in the room.
• Radiation: Some heat is emitted as infrared rays, though it's a smaller fraction in this scenario.

By understanding these processes, you can better predict temperature changes in environments like a classroom.

Specific Heat Capacity

Specific heat capacity indicates how much energy is needed to change the temperature of a substance. It is a key factor in determining temperature changes when heat is added or removed.
In our problem, the specific heat capacity (\[c = 1020 \, \text{J/(kg} \cdot \text{K)}\]) of air helps us calculate how much the temperature will rise when heat energy is absorbed. The key points are:

• A high specific heat capacity means the substance requires more energy for the same temperature change.
• For the lecture hall, this means the air can absorb a large amount of heat for a minimal temperature increase.

Specific heat capacity provides insight into why different materials and substances react differently to heat.

Energy Conversion

Energy conversion in thermodynamics involves changing energy from one form to another. In this scenario, the students' metabolic energy is converted to heat energy.
This conversion process occurs as:

• Students convert the energy from food into bodily functions, including generating heat.
• The heat energy is then released into the surrounding air.

Recognizing how energy conversion works helps explain the chain reaction of processes, making it easier to predict outcomes like temperature rises. Understanding energy conversions not only explains how heat affects environments, but also is crucial for practical applications like energy conservation.

Temperature Change

Temperature change results from the energy conversion and heat transfer between objects or substances. The formula for temperature change is:
\[\Delta T = \frac{Q}{m \cdot c}\]Where:

• \(Q\) is the total heat energy transferred.
• \(m\) is the mass of the air.
• \(c\) is the specific heat capacity of air.

Using this formula, we can calculate how much the classroom air's temperature rises due to students' heat output:
1. Listening students cause less temperature rise as they produce less heat.
2. Exam-taking students generate more heat, causing a greater temperature change.
Examining these results allows us to understand how controlled environments change with different internal conditions.