/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 . Hot air in a physics lecture. ... [FREE SOLUTION] | 91Ó°ÊÓ

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. Hot air in a physics lecture. (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 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 capacity 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 min in this case?

Short Answer

Expert verified
(a) 27,000,000 J; (b) 6.86 K; (c) 19.06 K

Step by step solution

01

Determine Total Heat Energy Output - (a)

Each student has a heat output of 100 watts, so a class of 90 students will have a cumulative heat output. Power is the rate of energy transfer, where 1 watt equals 1 joule per second. Therefore, the total energy output for 90 students in 50 minutes is:\[\text{Total Energy} = 90 \times 100 \times 3000\, \text{J} = 27,000,000\, \text{J}\] where 3000 is the total seconds in 50 minutes (50 \times 60).
02

Calculate Temperature Rise - (b)

Given the total heat energy from Step 1, we now consider it being absorbed by 3200 cubic meters of air. First, calculate the mass of the air using its density:\[\text{Mass of air} = 3200 \, \text{m}^3 \times 1.20 \, \text{kg/m}^3 = 3840 \, \text{kg}\]Using the formula for temperature rise \( \Delta T \), where \( Q = mc\Delta T \), solve for \( \Delta T \):- \( Q = 27,000,000 \, \text{J} \)- \( m = 3840 \, \text{kg} \)- \( c = 1020 \, \text{J/kg} \cdot \text{K} \)\[\Delta T = \frac{Q}{mc} = \frac{27,000,000}{3840 \times 1020} \approx 6.86 \, \text{K} \]
03

Determine Heat Output Increase for Exam - (c)

When the class is taking an exam, each student has a heat output of 280 watts. Calculate the new total energy output for 90 students in a 50-minute period:\[\text{Total Energy} = 90 \times 280 \times 3000 \, \text{J} = 75,600,000 \, \text{J}\]
04

Calculate New Temperature Rise with Increased Heat Output - (c)

Using the same mass of air and specific heat capacity, now solve for the new temperature rise \( \Delta T \) with the increased heat energy:- \( Q = 75,600,000 \, \text{J} \)\[\Delta T = \frac{75,600,000}{3840 \times 1020} \approx 19.06 \, \text{K}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
Heat transfer is the process by which heat energy moves from one place to another. In the context of a classroom, this means the energy generated by students' bodies is distributed throughout the room.
All forms of heat transfer require a difference in temperature and proceed according to the second law of thermodynamics, which states that heat flows from a warmer object to a cooler one.
In our classroom scenario, student bodies generate heat, raising the temperature of the surrounding air.

The three main modes of heat transfer are conduction, convection, and radiation. However, in large spaces like lecture halls, convection plays a significant role.
  • **Conduction**: Transfer of heat through direct contact, often negligible in large air volumes.
  • **Convection**: Circulating air carries heat away from its source, distributing it evenly.
  • **Radiation**: Heat energy radiates out without needing a medium. This can contribute to overall heat distribution in the room.
This energy exchange impacts the overall air temperature, especially when an air-conditioning system is not in use.
Specific Heat Capacity
Specific heat capacity is a property that determines how much heat energy a substance can store per unit of mass. It's represented by the symbol 'c' and is usually expressed in joules per kilogram per Kelvin (\(\mathrm{J}/(\mathrm{kg} \, \cdot \, \mathrm{K})\)).
A higher specific heat capacity means a substance can absorb more heat without a significant increase in temperature.

For air in the classroom, the specific heat capacity is 1020 \(\mathrm{J}/(\mathrm{kg} \, \cdot \, \mathrm{K})\).
This means air can absorb a substantial amount of heat before its temperature rises significantly.
  • This property is crucial in scenarios where air conditioning is turned off as it prevents rapid temperature rises.
  • Understanding specific heat helps in calculating how much energy is needed to cause a certain temperature change.
Earth's atmospheric conditions make air's specific heat capacity an essential factor in thermal energy transfers, impacting how quickly a room heats or cools.
Temperature Rise
Temperature rise occurs when heat energy is added to a substance, leading to an increase in its thermal energy.
In a closed space like a classroom, heat produced by students will raise the temperature of the room air. To quantify this rise, we use the formula:
\[\Delta T = \frac{Q}{mc}\]
Where:
  • \(\Delta T\) is the change in temperature
  • \(Q\) is the heat added (in joules)
  • \(m\) is the mass of the air (in kilograms)
  • \(c\) is the specific heat capacity of air
This equation helps us understand how much the temperature will increase when a specific amount of heat is applied.

For example, if 27,000,000 joules of heat energy is added to 3840 kg of air with a specific heat capacity of 1020 J/kg·K, the temperature rise can be calculated. This results in approximately a 6.86 K rise in temperature.
Energy Output Calculation
Energy output calculation is essential for understanding the total amount of energy produced over time. In our problem, each student generates heat at a rate of 100 W (watts), where 1 W equals 1 joule per second. Therefore, for a class of 90 students:
  • Calculate the total energy output for the duration of class:
    \[\text{Total Energy} = 90 \times 100 \times 3000 = 27,000,000 \, \text{J}\]
  • The number 3000 represents the total number of seconds in 50 minutes (50 \times 60).
When taking an exam, students' energy output increases to 280 W due to more intense cognitive work. Thus, the total energy for this situation becomes:
\[\text{Total Energy} = 90 \times 280 \times 3000 = 75,600,000 \, \text{J}\]
These calculations allow us to determine how much heat contributes to the rising temperature in a given environment, which is crucial for thermal management strategies, especially in settings lacking sufficient ventilation or cooling systems. By understanding energy output, educators can better prepare for the thermal dynamics of their classrooms, whether through scheduling, room design, or implementing cooling solutions.

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Most popular questions from this chapter

\(\cdot\) A nail driven into a board increases in temperature. If 60\(\%\) of the kinetic energy delivered by a 1.80 kg hammer with a speed of 7.80 \(\mathrm{m} / \mathrm{s}\) is transformed into heat that flows into the nail and does not flow out, what is the increase in temperature of an 8.00 g aluminum nail after it is struck 10 times?

\bullet A 15.0 g bullet traveling horizontally at 865 \(\mathrm{m} / \mathrm{s}\) passes through a tank containing 13.5 \(\mathrm{kg}\) of water and emerges with a speed of 534 \(\mathrm{m} / \mathrm{s}\) . What is the maximum temperature increase that the water could have as a result of this event?

Temperatures in biomedicine. (a) Normal body temperature. The average normal body temperature measured in the mouth is 310 \(\mathrm{K}\) . What would Celsius and Fahrenheit thermometers read for this temperature? (b) Elevated body temperature. During very vigorous exercise, the body's temperature can go as high as \(40^{\circ} \mathrm{C}\) . What would Kelvin and Fahrenheit thermometers read for this temperature? (c) Temperature difference in the body. The surface temperature of the body is normally about 7 \(\mathrm{C}^{\circ}\) lower than the internal temperature. Express this temperature difference in kelvins and in Fahrenheit degrees. (d) Blood storage. Blood stored at \(4.0^{\circ} \mathrm{C}\) lasts safely for about 3 weeks, whereas blood stored at \(-160^{\circ} \mathrm{C}\) lasts for 5 years. Express both temperatures on the Fahrenheit and Kelvin scales. (e) Heat stroke. If the body's temperature is above \(105^{\circ} \mathrm{F}\) for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

\(\bullet\) A copper pot with a mass of 0.500 \(\mathrm{kg}\) contains 0.170 \(\mathrm{kg}\) of water, and both are at a temperature of \(20.0^{\circ} \mathrm{C} . \mathrm{A} 0.250 \mathrm{kg}\) block of iron at \(85.0^{\circ} \mathrm{C}\) is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

A Styrofoam" bucket of negligible mass contains 1.75 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice. More ice, from a refrigerator at \(-15.0^{\circ} \mathrm{C},\) is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.778 \(\mathrm{kg} .\) Assuming no heat exchange with the surroundings, what mass of ice was added?

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