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Chapter 15: Problem 3

In moving out of a dormitory at the end of the semester, a student does \(1.6 \times 10^{4} \mathrm{~J}\) of work. In the process, his internal energy decreases by \(4.2 \times 10^{4} \mathrm{~J}\). Determine each of the following quantities (including the algebraic sign): (a) \(W,(\mathrm{~b}) \Delta U\) and \((\mathrm{c}) Q\)

Short Answer

Expert verified
(a) \(W = 1.6 \times 10^4 \) J, (b) \(\Delta U = -4.2 \times 10^4 \) J, (c) \(Q = -2.6 \times 10^4 \) J.

Step by step solution

01

Understanding Energy Changes

The problem involves the internal energy change (\(\Delta U\)) of the student. We know that the work done by the student is \(W = 1.6 \times 10^4\, \text{J}\). His internal energy decreases, meaning \(\Delta U = -4.2 \times 10^4\, \text{J}\).
02

Applying the First Law of Thermodynamics

The First Law of Thermodynamics states \( \Delta U = Q - W \), where \(Q\) is the heat exchanged, \(\Delta U\) is the change in internal energy, and \(W\) is the work done by the system. We can rearrange this equation to solve for \(Q\): \(Q = \Delta U + W\).
03

Substituting Known Values to Find Q

Substitute \(\Delta U = -4.2 \times 10^4 \) J and \(W = 1.6 \times 10^4 \) J into the equation \(Q = \Delta U + W\): \[ Q = (-4.2 \times 10^4) + (1.6 \times 10^4) \].
04

Calculating Q

Perform the calculation: \[ Q = -4.2 \times 10^4 + 1.6 \times 10^4 = -2.6 \times 10^4 \text{ J} \]. This negative value indicates that the student lost heat.

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Key Concepts

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

Understanding Internal Energy
Internal energy is a key concept in thermodynamics tailored for understanding energy within a system. Think of it as the total of all molecular kinetic and potential energies stored in a system. Changes in internal energy occur due to heat transfer or work being done on or by the system.
In our exercise, the student's internal energy decreases by \(-4.2 \times 10^4\; \text{J}\). This negative change signals energy lost by the body, driven by processes like doing physical work, which diverts energy into moving objects.
Even without focusing on chemical intricacies, the takeaway is simple: when energy leaves or enters a system, its internal energy changes. This is fundamental in figuring out transformations in energetic states, much like what the student experiences when exerting physical effort.
The Concept of Work Done
Work done involves energy transferred by movement against a force. In physical terms, it means exerting force over a distance, like lifting a box or moving furniture.
In our example, the student did \(1.6 \times 10^4\; \text{J}\) of work. This involves applying force, putting effort into activity, and moving objects out of the dorm. Whenever you do work like this, energy is spent, impacting internal reserves.
  • Positive Work: When work is performed or effort is given by the system.
  • Negative Work: When work is done on the system.
Here, the system鈥攐ur student鈥攄elivers energy through work done, reflecting a positive value. This is part of the energy balance dictated by the First Law of Thermodynamics.
Understanding Heat Exchanged
Heat exchanged is the energy transfer due to a temperature difference. It flows from warmer to cooler areas until thermal equilibrium is achieved.
In our problem, we find heat exchanged using the First Law of Thermodynamics: \( \Delta U = Q - W \). By substituting the known values: \( \Delta U = -4.2 \times 10^4 \; \text{J} \) and \( W = 1.6 \times 10^4 \; \text{J} \), we calculate \( Q \): \[ Q = (-4.2 \times 10^4) + (1.6 \times 10^4) = -2.6 \times 10^4 \; \text{J} \]. This negative \( Q \) value means heat was lost during this process.
  • Positive Q: Heat absorbed by the system.
  • Negative Q: Heat emitted by the system.
In essence, energy transformation isn't just about work but also involves the intricate heat transactions that define state changes, either heating up or cooling down, as demonstrated by our student's experience.

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Most popular questions from this chapter

Concept Questions An ideal gas expands isothermally, doing work, while heat flows into the gas. (a) Does the internal energy of the gas increase, decrease, or remain the same? (b) Is the work done by the gas greater than, less than, or equal to the heat that flows into the gas? Account for your answers. Problem Three moles of neon expand isothermally from 0.100 to \(0.250 \mathrm{~m}^{3}\). Into the gas flows \(4.75 \times 10^{3} \mathrm{~J}\) of heat. Assuming that neon is an ideal gas, find its temperature.

The drawing (not to scale) shows the way in which the pressure and volume change for an ideal gas that is used as the working substance in a Carnot engine. The gas begins at point a (pressure \(=P_{\mathrm{a}}\), volume \(\left.=V_{\mathrm{a}}\right)\) and expands isothermally at temperature \(T_{\mathrm{H}}\) until point \(\mathrm{b}\) (pressure \(=P_{\mathrm{b}}\), volume \(=V_{\mathrm{b}}\) ) is reached. During this expansion, the input heat of magnitude \(\left|Q_{\mathrm{H}}\right|\) enters the gas from the hot reservoir of the engine. Then, from point \(\mathrm{b}\) to point \(\mathrm{c}\) (pressure \(=P_{\mathrm{c}}\), volume \(\left.=V_{\mathrm{c}}\right)\), the gas expands adiabatically. Next, the gas is compressed isothermally at temperature \(T_{\mathrm{C}}\) from point \(\mathrm{c}\) to point \(\mathrm{d}\) (pressure \(=P_{\mathrm{d}}\), volume \(=V_{\mathrm{d}}\) ). During this compression, heat of magnitude \(\left|Q_{\mathrm{C}}\right|\) is rejected to the cold reservoir of the engine. Finally, the gas is compressed adiabatically from point \(d\) to point a, where the gas is back in its initial state. The overall process a to \(b\) to \(\mathrm{c}\) to \(\mathrm{d}\) is called a Carnot cycle. Prove for this cycle that \(|Q \mathrm{c}| /|Q \mathrm{H}|=T_{\mathrm{C}} / T_{\mathrm{H}}\)

Concept Questions Two Carnot air conditioners, \(\mathrm{A}\) and \(\mathrm{B},\) are removing heat from different rooms. The outside temperature is the same for both, but the room temperatures are different. The room serviced by unit \(\mathrm{A}\) is kept colder than the room serviced by unit B. The heat removed from both rooms is the same, (a) Which unit requires the greater amount of work? (b) Which unit deposits the greater amount of heat outside? Explain. Problem The outside temperature is \(309.0 \mathrm{~K}\). The room serviced by unit \(\mathrm{A}\) is kept at a temperature of \(294.0 \mathrm{~K},\) while the room serviced by unit \(\mathrm{B}\) is kept at \(301.0 \mathrm{~K}\). The heat removed from either room is \(4330 \mathrm{~J} .\) For both units, find the magnitude of the work required and the magnitude of the heat deposited outside. Verify that your answers are consistent with your answers to the Concept Questions.

The temperature of \(2.5 \mathrm{~mol}\) of a monatomic ideal gas is \(350 \mathrm{~K}\). The internal energy of this gas is doubled by the addition of heat. How much heat is needed when it is added at (a) constant volume and (b) constant pressure?

A system does \(4.8 \times 10^{4} \mathrm{~J}\) of work, and \(7.6 \times 10^{4} \mathrm{~J}\) of heat flows into the system in the process. (a) Considered by itself, does the work increase or decrease the internal energy of the system? (b) Considered by itself, does the heat increase or decrease the internal energy? (c) Considering the work and heat together, does the internal energy of the system increase, decrease, or remain the same? Explain Find the change in the internal energy of the system. Make sure that your answer is consistent with your answers to the Concept Questions.
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