/*! 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 18 A student performs a combustion ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A student performs a combustion experiment by burning a mixture of fuel and oxygen in a constant-volume metal can surrounded by a water bath. During the experiment the temperature of the water is observed to rise. Regard the mixture of fuel and oxy gen as the system. (a) Has heat been transferred? How can you tell? (b) Has work been done? How can you tell? (c) What is the sign of \(\Delta U ?\) How can you tell?

Short Answer

Expert verified
Yes, heat is transferred; no work is done; \(\Delta U < 0\).

Step by step solution

01

Understanding Heat Transfer

In any combustion reaction, energy is released in the form of heat. Since the temperature of the water bath increases, it indicates that heat has been transferred to the surrounding water from the system (the mixture of fuel and oxygen). Therefore, there is heat transfer from the system to the surroundings, which means heat has definitely been transferred. This is observed as the increase in the water temperature.
02

Determining if Work is Done

Work is done by a system when there is a volume change. In this scenario, the combustion occurs in a constant-volume metal can, meaning no physical expansion occurs. Therefore, no work is done by the system on the surroundings because the volume remains constant.
03

Analyzing the Sign of Internal Energy Change ( \Delta U )

The first law of thermodynamics states that the change in internal energy \(\Delta U\) is equal to heat \(q\) transferred into the system minus work \(w\) done by the system, \(\Delta U = q - w\). Since the system releases heat to the water bath, \(q\) is negative, and since no work is done, \(w = 0\). Thus, \(\Delta U = q - 0 < 0\), which means the change in internal energy of the system is negative.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics and plays a crucial role in many physical processes. In the context of the combustion experiment described, heat transfer occurs when the energy from the burning mixture of fuel and oxygen moves to the surrounding water bath. This movement of thermal energy results in an increase in the water's temperature, signaling that heat has been successfully transferred.
  • Heat is defined as the energy in transit due to temperature differences between the system and its surroundings.
  • In this experiment, the heat is transferred from the system (the mixture) to the water bath, illustrating how heat naturally moves from hotter to cooler objects.
Understanding heat transfer is essential for explaining temperature changes observed in experiments and various natural processes. It provides insight into why certain systems gain or lose energy and how they interact with their surroundings.
First Law of Thermodynamics
The First Law of Thermodynamics is a critical principle in understanding energy conservation. It states that the total energy in a closed system remains constant—energy can neither be created nor destroyed, only transformed from one form to another. In the combustion experiment, this law helps us predict the changes in internal energy, heat transfer, and work done. The law formulates that the change in internal energy \(\Delta U\) of a system is equal to the heat \(q\) added to the system minus the work \(w\) done by the system on its surroundings:\[ \Delta U = q - w \]
  • For the experiment, since no work is done (as volume is constant), we focus on heat transfer.
  • The heat released from the system is negative (\(q < 0\)), leading to a decrease in internal energy (\(\Delta U < 0\)).
By applying this first law, one better grasps the balance of energy transactions in physical systems.
Internal Energy
Internal energy is the total energy contained within a system, encompassing all kinetic and potential energies of the molecules. During the combustion process in the experiment, the internal energy decreases as energy is released in the form of heat.The change in internal energy \(\Delta U\) is crucial for understanding how systems exchange energy with their surroundings:
  • If energy is released by the system as heat, the internal energy decreases.
  • The experiment's system (fuel and oxygen mixture) loses energy to the surrounding water, resulting in negative \(\Delta U\).
In thermodynamics, analyzing \(\Delta U\) helps explain why systems experience energy shifts, such as cooling down or heating up, due to energy interactions with external environments. Knowing \(\Delta U\) assists in making predictions about how a system behaves under various conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two moles of carbon monoxide (CO) start at a pressure of 1.2 atm and a volume of 30 liters. The gas is then compressed adiabatically to \(\frac{1}{3}\) this volume. Assume that the gas may be treated as ideal. What is the change in the internal energy of the gas? Does the internal energy increase or decrease? Does the temperature of the gas increase or decrease during this process? Explain.

In a cylinder sealed with a piston, you rapidly compress 3.00 \(\mathrm{L}\) of \(\mathrm{N}_{2}\) gas initially at 1.00 atm pressure and \(0.00^{\circ} \mathrm{C}\) to half its original volume. Assume the \(\mathrm{N}_{2}\) behaves like an ideal gas. (a) Calculate the final temperature and pressure of the gas. (b) If you now cool the gas back to \(0.00^{\circ} \mathrm{C}\) without changing the pressure, what is its final volume?

Two moles of an ideal gas are compressed in a cylinder at a constant temperature of \(85.0^{\circ} \mathrm{C}\) until the original pressure has tripled. (a) Sketch a \(p V\) -diagram for this process. (b) Calculate the amount of work done.

Five moles of an ideal monatomic gas with an initial temperature of \(127^{\circ} \mathrm{C}\) expand and, in the process, absorb 1200 \(\mathrm{J}\) of heat and do 2100 \(\mathrm{J}\) of work. What is the final temperature of the gas?

A cylinder contains 0.0100 mol of helium at \(T=27.0^{\circ} \mathrm{C}\) (a) How much heat is needed to raise the temperature to \(67.0^{\circ} \mathrm{C}\) while keeping the volume constant? Draw a \(p V\) -diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from \(27.0^{\circ} \mathrm{C}\) to \(67.0^{\circ} \mathrm{C}\) ? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? d) If the gas is ideal, what is the change in its internal energy in part (a)? In part \((b) ?\) How do the two answers compare? Why?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.