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What is the change in internal energy of a system which does 4.50×105J of work while 3.00×106J of heat transfer occurs into the system, and 8.00×106J of heat transfer occurs to the environment?

Short Answer

Expert verified
The change in internal energy of the system is \( -5.45 \times 10^6\,J \).

Step by step solution

01

Understand the First Law of Thermodynamics

The change in internal energy of a system can be determined using the first law of thermodynamics, which is stated as \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy of the system, \( Q \) is the net heat transfer into the system, and \( W \) is the work done by the system.
02

Calculate the Net Heat Transfer (Q)

The net heat transfer is the difference between the heat transferred into the system and the heat transferred to the environment. In this case, it is \( Q = Q_{in} - Q_{out} = 3.00 \times 10^6\,J - 8.00 \times 10^6\,J = -5.00 \times 10^6\,J \).
03

Calculate the Change in Internal Energy (Delta U)

Using the first law of thermodynamics and substituting the values, we get \( \Delta U = -5.00 \times 10^6\,J - 4.50 \times 10^5\,J = -5.45 \times 10^6\,J \). Therefore, the change in internal energy of the system is \( -5.45 \times 10^6\,J \).

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Key Concepts

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

Internal Energy
When discussing the mysteries of thermodynamics, we often talk about a concept known as internal energy. This is a measure of the energy stored within a system, typically attributed to the kinetic and potential energy of the particles within. Think of it like the hidden power reserve that can't be seen with the naked eye, but is definitely there, much like the battery in a smartphone.

Delving into our exercise, when work is executed by a system, or heat is transferred into, or out of it, the internal energy changes. Remember, this change is denoted by \( \Delta U \), and it reflects the difference in the energy from before to after an event. In our case, performing work and exchanging heat with the environment are the key players in the game of energy change.

Vital to remember is the sign convention for internal energy: Positive \( \Delta U \) occurs when a system gains energy, while a negative \( \Delta U \) means the system has lost energy. So, what does our negative solution for \( \Delta U \) tell us? Simply put, our system gave out more energy than it received, much like spending more money than you earn. It's an energy deficit.
Heat Transfer
Stepping into the world of heat transfer, a topic sizzling with action, it's the process of energy flowing from a warmer place to a cooler one. But in thermodynamics, we give this process a numeric value, which helps us quantify how much energy is getting cozy in one system versus taking an exit to another.

In the given exercise, heat transfer played a pivotal role. There were two types to consider: the heat coming into the system \( Q_{in} \) and the heat bowing its way out \( Q_{out} \) to the environment. By finding the difference between these two (\( Q_{in} - Q_{out} \)), we're able to capture the net heat transfer (\( Q \) in our formula).

Always keep an eye on the direction of heat flow. Is it incoming or outgoing? It will guide the signs you use. Heat entering the system is usually considered positive, and when it leaves, it's negative. Our negative \( Q \) suggested that our system wasn't just chilling; it was quite literally chilling, shedding more energy than it absorbed.
Thermodynamic Work
When we chat about thermodynamic work, we're talking about the process of energy being used to perform tasks—it could be compression, expansion, lifting, or any mechanical action. One thing to note is that work can be done by the system or on the system, which will affect the energy balance accordingly.

According to the rules of the thermodynamics playbook, if a system does work, like in our exercise, pushing the barriers or moving objects, this work is considered positive. And, as you've likely guessed, that means the internal energy decreases, since the system is using up energy to accomplish this work.

But here's a hook: if work is done on the system (imagine compressing a piston), then the work value is negative, and the system's internal energy goes up. It’s taking energy in, not giving it away. In the case of our exercise, the work \( W \) done by the system carried a positive sign, and the internal energy, taking a bow, stepped down by the same amount.

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Most popular questions from this chapter

Assume that the turbines at a coal-powered power plant were upgraded, resulting in an improvement in efficiency of 3.32%. Assume that prior to the upgrade the power station had an efficiency of 36% and that the heat transfer into the engine in one day is still the same at 2.50×1014J. (a) How much more electrical energy is produced due to the upgrade? (b) How much less heat transfer occurs to the environment due to the upgrade?

A rotating space station is said to create "artificial gravity" - a loosely- defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an "artificial gravity" of \(9.80 \mathrm{m} / \mathrm{s}^{2}\) at the rim?

Heat transfer Q and work done W are always energy in transit, whereas internal energy U is energy stored in a system. Give an example of each type of energy, and state specifically how it is either in transit or resides in a system.

Modern roller coasters have vertical loops like the one shown in Figure \(6.38 .\) The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \(15.0 \mathrm{m}\) and the downward acceleration of the car is \(1.50 \mathrm{g} ?\) Figure 6.38 Teardrop-shaped loops are used in the latest roller coasters so that the radius of curvature gradually decreases to a minimum at the top. This means that the centripetal acceleration builds from zero to a maximum at the top and gradually decreases again. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered long ago in railroad curve design. With a small radius of curvature at the top, the centripetal acceleration can more easily be kept greater than \(g\) so that the passengers do not lose contact with their seats nor do they need seat belts to keep them in place.

What is the change in entropy in an adiabatic process? Does this imply that adiabatic processes are reversible? Can a process be precisely adiabatic for a macroscopic system?

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