How Does Running Two Heat Engines Affect the Second Law of Thermodynamics?

In summary, the firebox is at 750K and the ambient temp is 300K. The efficiency of a Carnot engine doing 150J of work as it transports energy between these constant temperature baths is 60%. The Carnot engine must take in energy 150/.60=250J from the hot reservoir and must put out 100J of energy by heat into the environment. To follow Carnot's reasoning, suppose that xome other heat engine S could have efficiency 70%. If engine S operated as in part a) and ran the Carnot engine in reverse, then the total energy the firebox puts out as both engines operate together would be -35.7J. However, the Clausius statement of the second law of thermodynamics
  • #1
Erik Horwath
7
0
a firebox is at 750K and the ambient temp is 300K. The efficiency of a Carnot engine doing 150J of work as it transports energy between these constant temperature baths is 60%. The Carnot engine must take in energy 150/.60=250J from the hot reservoir and must put out 100J of energy by heat into the environment. To follow Carnot's reasoning, suppose that xome other heat engine S could have efficiency 70%. a)Find the energy input and wasted energy output of engine S as it does 150J of work. b)Let engine S operate as in part a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statemetn of the second law of thermodynamics is violated.

Part a) is obvious. They show exactly how to do it within the problem. But the wording in part b) is confusing to me. If the engine is operating in reverse, then it is a heat pump, not a heat engine. Therefore the work under consideration must be input work, not output work. So when it says "let engine S operate as in part a)" does that mean 150J of work in done on the system rather than by the systm? Further, the expression for the coefficient of performance of a heat pump is different from that for the effiecency of a heat engine. The former is e=Wnet/Qin and is always less than unity and the latter is cop=Qout/Win and is generally >1. I don't understand what the problem is asking exactly. Thanks for your help! :smile:
 
Physics news on Phys.org
  • #2
You should show that the whole heat is turned into work.
 
  • #3
Maxos said:
You should show that the whole heat is turned into work.

I'm still confused as to how to proceed. I think you are confusing the Clausius statement of the second law (which essentially says that the spontaneous transfer of heat does not occur from a cold reservoir to a hot reservoir) with the Kelvin-Planck statement (which says that it is not possible to design a perfectly efficient heat engine - i.e. one that converts all input heat to work). The problem asks me to show that running two heat engines simultaneously forward and in reverse (between the same two energy reservoirs) results in the transfer of heat from a cold reservoir to a hot reservoir without the positive input work.

The question is an odd - the answer in the back of the text is: "-35.7J, -35.7J. The net effect is the transport of energy by heat from the cold to the hot reservoir without the expenditure of work."
 

Related to How Does Running Two Heat Engines Affect the Second Law of Thermodynamics?

1. What is a Carnot heat engine?

A Carnot heat engine is a theoretical thermodynamic system that operates between two heat reservoirs at different temperatures and is able to convert heat energy into mechanical work. It follows the Carnot cycle, which consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.

2. How does a Carnot heat engine work?

A Carnot heat engine operates by absorbing heat energy from a high temperature reservoir, converting some of it into mechanical work, and releasing the remaining heat energy into a low temperature reservoir. This process is repeated in a cycle to produce continuous work output.

3. What is the Carnot efficiency?

The Carnot efficiency is the maximum efficiency that a heat engine can achieve, and it is equal to the ratio of the temperature difference between the two reservoirs to the high temperature reservoir's temperature. This efficiency is only achievable in a reversible Carnot cycle.

4. What is the Carnot heat engine problem?

The Carnot heat engine problem is a theoretical problem that asks whether it is possible to construct a heat engine with 100% efficiency, meaning all of the heat energy absorbed is converted into work without any energy loss. The Carnot heat engine is the solution to this problem, but it is not achievable in practice due to various factors such as friction and heat loss.

5. What are the real-life applications of the Carnot heat engine?

The Carnot heat engine has several real-life applications, such as in power plants, refrigerators, and air conditioners. It is also used as a model to analyze and improve the efficiency of real heat engines, and as a theoretical benchmark for comparing the performance of different engines.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
785
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Replies
1
Views
844
Replies
29
Views
7K
  • Introductory Physics Homework Help
Replies
11
Views
2K
Replies
6
Views
12K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Replies
6
Views
2K
Back
Top