Problem 97 - University Physics with Modern Physics

Question

Hot Air in a Physics Lecture. (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 specific heat 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and density 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 min in this case?

Step-by-Step Solution

Answer

The temperature rise in part (a) is 6.82 K, and in part (c) it's 19.09 K.

Step 1: Calculate Total Heat Output for Part (a)

First, find the total heat output of the class. Each student outputs 100 W. Therefore, a class of 90 students outputs 90 × 100 = 9000 W. Over a 50-minute lecture, this equates to energy: \(9000 \text{ W} \times 3000 \text{ seconds} = 27,000,000 \text{ J}\).

Step 2: Calculate Heat Capacity for Part (b)

The air in the room has a volume of 3200 m³, a density of 1.20 kg/m³, and a specific heat of 1020 J/(kg·K). First, calculate the mass of the air: \(3200 \text{ m}^3 \times 1.20 \text{ kg/m}^3 = 3840 \text{ kg}\).

Step 3: Calculate Temperature Increase for Part (b)

Using the formula \(Q = mc\Delta T\), rearrange to find \(\Delta T\): \(\Delta T = \frac{Q}{mc}\). Substitute \(Q = 27,000,000 \text{ J}\), \(m = 3840 \text{ kg}\), \(c = 1020 \text{ J/(kg·K)}\). Thus, \(\Delta T = \frac{27,000,000}{3840 \times 1020} \approx 6.82 \text{ K}\).

Step 4: Calculate New Heat Output for Part (c)

With students taking an exam, the heat output rises to 280 W per student. For 90 students, the total heat output is \(90 \times 280 = 25,200 \text{ W}\). Over 50 minutes, this is \(25,200 \text{ W} \times 3000 \text{ seconds} = 75,600,000 \text{ J}\).

Step 5: Calculate New Temperature Rise for Part (c)

Using \(Q = mc\Delta T\) again, find \(\Delta T\) using the new \(Q = 75,600,000 \text{ J}\): \(\Delta T = \frac{75,600,000}{3840 \times 1020} \approx 19.09 \text{ K}\).

Key Concepts

Heat Transfer

Heat transfer is the process of energy moving from one body or material to another. In the context of our physics lecture, students are generating heat that spreads throughout the lecture hall.
Heat can transfer in different ways:

• Conduction: Direct contact transfer of heat. Not relevant here as air is the medium.
• Convection: Movement of heat through fluid motion, which is significant in this scenario.
• Radiation: Heat transfer through electromagnetic waves.

In this problem, heat energy radiated by students is mainly transferred to the air through convection and radiated waves.

Specific Heat

Specific heat is a measure of how much heat energy a substance can hold before its temperature changes. It is crucial in determining how much the temperature of the air in the lecture hall rises.
We use the formula:\[Q = mc\Delta T\]where:

• \(Q\) is the heat energy input or output
• \(m\) is the mass of the substance (in kilograms)
• \(c\) is the specific heat capacity (J/kg·K)
• \(\Delta T\) is the change in temperature (in Kelvin)

In this problem, the specific heat of air is given as 1020 J/(kg·K), which helps calculate how much energy is needed to change the air temperature in the lecture hall.

Temperature Change

Understanding temperature change is essential to solving heat transfer problems. The temperature change indicates how a system's thermal state alters due to energy input.
To find the temperature change (\(\Delta T\)), we rearrange the heat formula:\[\Delta T = \frac{Q}{mc}\]From the example, when students release energy into the room, this value demonstrates how much the air's temperature would increase over time, providing real-world implications for comfort and climate control in indoor environments.

Energy Conversion

Energy conversion refers to changing energy from one form to another. In our physics problem, students convert metabolic energy into heat. This heat then affects the air's temperature.
Key concepts:

• From metabolic processes, chemical energy is converted to thermal energy.
• This thermal energy is measured in joules (J), the SI unit for energy.

The understanding of energy conversion in this scenario allows us to calculate how energy from the students' bodies impacts the room’s environment, which ties into thermodynamic principles.

Physics Lecture Problems

Physics lecture problems are designed to incorporate real-world applications of physical laws and principles. They help students understand how theoretical concepts apply in practical scenarios.
This exercise with the heat transfer in a lecture hall demonstrates:

• Application of thermodynamic equations for calculating energy flow.
• Real-life implications of physics, like room temperature changes due to human metabolism.
• Critical thinking and problem-solving using scientific methods.

By solving these problems, students gain a deeper understanding of physical phenomena and improve their analytical skills.