Problem 87 - University Physics with Modern Physics

Question

(a) A typical student listening attentively to a physics lecture has a heat output of 100 W. How much heat energy does a class of 140 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 m\(^3\) of air in the room. The air has specific heat 1020 J/kg \(\cdot\) K and density 1.20 kg/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 W. What is the temperature rise during 50 min in this case?

Step-by-Step Solution

Answer

(a) 42,000,000 J, (b) 10.54 K, (c) 29.61 K increase in temperature.

Step 1: Calculate Energy Output for Part (a)

To find the total heat energy released by 140 students, we start by calculating the energy output of one student. Each student releases 100 W (watts) of heat. Over the course of a 50-minute lecture, each student releases energy given by: Energy per student = Power \( \times \) Time = \( 100 \text{ W} \times 50 \times 60 \text{ s} = 300,000 \text{ J (joules)} \). For 140 students, the total energy output is: Total energy =\( 140 \times 300,000 \text{ J} = 42,000,000 \text{ J} \).

Step 2: Calculate Temperature Rise for Part (b)

Next, we need to find out how much the temperature of the air rises when this energy is absorbed by the air in the room. First, calculate the mass of the air in the room using the formula: \( \text{Mass} = \text{Density} \times \text{Volume} = 1.20 \text{ kg/m}^3 \times 3200 \text{ m}^3 = 3840 \text{ kg} \) The specific heat capacity of air is given as 1020 J/kg \( \cdot \) K. The temperature change \( \Delta T \) can be found using the formula: \( \text{Energy} = \text{Mass} \times \text{Specific Heat} \times \Delta T \) Plug in the known values: \( 42,000,000 \text{ J} = 3840 \text{ kg} \times 1020 \text{ J/kg} \cdot \text{K} \times \Delta T \)Solve for \( \Delta T \): \( \Delta T = \frac{42,000,000 \text{ J}}{3840 \text{ kg} \times 1020 \text{ J/kg} \cdot \text{K}} \approx 10.54 \text{ K} \)

Step 3: Calculate Energy Output for Part (c)

Now, assume each student outputs 280 W during an exam. Energy per student = Power \( \times \) Time = \( 280 \text{ W} \times 50 \times 60 \text{ s} = 840,000 \text{ J} \).For 140 students, the total energy output is:Total energy = \( 140 \times 840,000 \text{ J} = 117,600,000 \text{ J} \).

Step 4: Calculate Temperature Rise for Part (c)

Using the same mass and specific heat calculations as in Step 2, determine the temperature rise with the increased energy: Plug in the known values for the new energy scenario:\( 117,600,000 \text{ J} = 3840 \text{ kg} \times 1020 \text{ J/kg} \cdot \text{K} \times \Delta T \)Solve for \( \Delta T \): \( \Delta T = \frac{117,600,000 \text{ J}}{3840 \text{ kg} \times 1020 \text{ J/kg} \cdot \text{K}} \approx 29.61 \text{ K} \)

Key Concepts

Specific Heat

When discussing heat transfer, the term "specific heat" plays a vital role. Specific heat is the amount of energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one kelvin). It is measured in joules per kilogram per kelvin (J/kg·K). In the context of our problem, the specific heat of air is given as 1020 J/kg·K. This value helps us determine how much energy is needed to increase the temperature of the air in the lecture hall.

In essence, specific heat can be thought of as a substance's resistance to temperature change. For example, water has a high specific heat, meaning it takes a lot of energy to change its temperature, whereas metals typically have lower specific heats. When solving problems, knowing the specific heat of the substances involved allows us to predict how much temperature change will occur given a certain amount of energy.

In our physics problem, the specific heat formula is used to calculate the rise in air temperature in the lecture hall given the energy output of the students. The formula used is:

• Specific Heat Formula: \( \text{Energy} = \text{Mass} \times \text{Specific Heat} \times \Delta T \)
• Where \( \Delta T \) represents the change in temperature.

By rearranging this formula to solve for \( \Delta T \), we can find out how much the temperature of the air in the room will increase after it absorbs a specific amount of heat energy.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed but only transferred or transformed from one form to another. This fundamental principle is crucial in physical calculations concerning heat transfer, such as the one in our problem.

All the heat produced by the students during the lecture is transferred to the air in the lecture hall. The energy conservation principle ensures us that the total amount of energy before and after the heat transfer remains constant. Understanding this allows us to predict outcomes like the temperature change by calculating quantities like energy and mass and equating them across different forms.

In the scenario provided, we relied on conservation of energy to calculate the temperature increases. All heat energy output by the students during the class and exam scenarios is transferred fantastically to the air because of no escape routes for heat—air conditioning systems are turned off, and it's presumed that the heat energy does not escape the lecture hall. These conditions simplify the problem, allowing direct application of energy conservation to analyze the situation. This assumption helped us to use the energy gained by the air to compute the temperature increase.

Physics Problem Solving

Tackling physics problems involves a systematic approach to ensure accuracy and clarity. The problem-solving process can be broken down into several key steps, enabling us to handle complex scenarios effectively.

Firstly, it's crucial to understand the problem by identifying known variables and what needs to be solved. In our exercise, the variables such as the power output per student, the total number of students, the air's specific heat capacity, and density were given upfront. Recognizing these allows us to construct a framework for further analysis.

Once we've organized our given information, the next step is to apply relevant physical laws and formulas. As seen in our problem:

• Energetic calculations were made using: \( \text{Energy} = \text{Power} \times \text{Time} \)
• For temperature change, we applied specific heat capacity theory with: \( \Delta T = \frac{\text{Energy}}{\text{Mass} \times \text{Specific Heat}} \)

Breaking complex equations into manageable parts—like separate calculations for mass and energy—is part of effective problem-solving.

Finally, evaluate and verify your results. Does the calculated outcome make sense in the context of the problem? Our calculations should reveal insightful answers, such as an understandable increase in room temperature when students' activity levels change. Confirming results with logical reasoning helps reinforce our understanding and proficiency in physics problem solving.