Problem 99 - University Physics with Modern Physics

Question

Hot Air in a Physics Lecture. (a) Atypical 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 capacity 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 \(\mathrm{min}\) in this case?

Step-by-Step Solution

Answer

(a) 27,000,000 J; (b) 6.84 K rise; (c) 19.23 K rise.

Step 1: Calculate Total Heat Output

First, calculate the total heat energy released by the students during the lecture. Each student produces 100 W of heat output. With 90 students present, the total heat output is \[\text{Total power} = 90 \times 100 \text{ W} = 9000 \text{ W}\]The lecture lasts 50 minutes, which is equivalent to 3000 seconds. So, the total heat energy released is given by the product of total power and time:\[\text{Total energy} = 9000 \text{ W} \times 3000 \text{ s} = 27,000,000 \text{ J}\]

Step 2: Calculate Temperature Rise of Air for Lecture

Now, calculate the temperature rise in the room assuming all the heat energy is transferred to the air. First, determine the mass of the air in the room using the density and the volume:\[\text{Mass of air} = \text{density} \times \text{volume} = 1.20 \text{ kg/m}^3 \times 3200 \text{ m}^3 = 3840 \text{ kg}\]The temperature rise \(\Delta T\) can be found using the specific heat capacity formula: \[Q = mc\Delta T\]Solving for \(\Delta T\):\[\Delta T = \frac{Q}{mc} = \frac{27,000,000 \text{ J}}{3840 \text{ kg} \times 1020 \text{ J/kg} \cdot \text{K}} = 6.84 \text{ K}\]

Step 3: Calculate Total Heat Output During Exam

If each student has a heat output of 280 W while taking an exam:\[\text{Total power} = 90 \times 280 \text{ W} = 25200 \text{ W}\]Over 3000 seconds (50 minutes):\[\text{Total energy} = 25200 \text{ W} \times 3000 \text{ s} = 75,600,000 \text{ J}\]

Step 4: Calculate Temperature Rise of Air During Exam

Using the same mass and specific heat capacity of air as before, calculate the temperature rise during the exam:\[\Delta T = \frac{Q}{mc} = \frac{75,600,000 \text{ J}}{3840 \text{ kg} \times 1020 \text{ J/kg} \cdot \text{K}} = 19.23 \text{ K}\]

Key Concepts

Heat Transfer

Heat transfer is an essential concept in thermodynamics, describing the movement of thermal energy from one object or substance to another due to temperature differences. This transfer can occur in three main ways: conduction, convection, and radiation.
Conduction occurs through direct contact where heat travels through a material without the material itself moving. It's like when a hot pot handle becomes warm even though it wasn't directly exposed to heat. Conversely, convection involves the transfer of heat through the movement of fluids or gases. You can imagine this when hot air rises and cool air sinks, creating a cycle of circulation. Finally, radiation does not require a medium; it transfers heat through electromagnetic waves, like the sun warming your face.
In the context of our problem, the heat output from students in a lecture hall is an example of convection. This heat energy from the students is transferred to the surrounding air, increasing its temperature. Understanding this concept helps clarify how heat from various sources can influence the environment, making it a vital subject in studying energy systems.

Specific Heat Capacity

Specific heat capacity is defined as the amount of heat energy required to raise the temperature of a given mass of a substance by one degree Celsius or Kelvin. This property varies among different substances, affecting how they heat up or cool down.
Mathematically, specific heat capacity is part of the equation:\[Q = mc\Delta T\]Where:

• \( Q \) is the total heat energy absorbed or released
• \( m \) is the mass of the substance
• \( c \) is the specific heat capacity
• \( \Delta T \) is the change in temperature

In this exercise, the specific heat capacity of air is used to determine how the heat output from students affects the temperature of the room. With a higher specific heat capacity, the air can absorb more heat without a significant rise in temperature, which explains why rooms do not heat up rapidly despite the energy produced by people inside them.
Understanding specific heat capacity is crucial as it allows you to predict how different materials or substances will respond to heat, which is critical in areas like cooking, climate control, and engineering.

Temperature Change

Temperature change refers to the difference between the initial and final temperatures of a system after heat is added or removed. Monitoring temperature changes is a common part of many scientific and engineering calculations, showing how systems respond to thermal inputs or outputs.
Using the formula mentioned earlier, \[\Delta T = \frac{Q}{mc}\]we can see how certain factors influence temperature change:

• If the amount of heat energy \( Q \) increases, the change in temperature \( \Delta T \) will also increase, given constant mass and specific heat capacity.
• Larger mass \( m \) or higher specific heat capacity \( c \) tends to moderate the temperature change, even with the same amount of heat energy added.

In our classroom scenario, the calculation of temperature change is essential. With students emitting heat, the classroom temperature will rise depending on the air's mass and specific heat capacity. During exams, when students generate more heat, there's a larger temperature increase, clearly demonstrating the relationship between heat output and temperature change.
This concept is foundational in managing thermal environments, be it in designing climate control systems or understanding natural climatic variations.