A heat engine and determining Power

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Homework Help Overview

The discussion revolves around a heat engine operating between two temperatures and its ability to lift a mass at a constant speed. The problem involves determining the power required by the engine based on the given parameters.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate temperature to work and power but is unsure of the equations needed. They express a need for guidance on calculating work and subsequently power. Other participants suggest focusing on the work required to lift the mass and how that relates to power output.

Discussion Status

Participants are exploring the relationship between work, power, and the parameters given in the problem. Some guidance has been offered regarding the calculation of work needed to lift the mass, but there is no explicit consensus on the overall approach to the problem.

Contextual Notes

There is mention of the operating temperatures of the heat engine, suggesting that additional factors may be relevant to the problem, though these have not been fully explored in the discussion.

JasonL
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A heat engine operating between temperatures 500K and 300K is used to lift a 10KG mass vertically at a constant speed of 4 meters per second.

Determine the power the engine must supply to life the mass.


I'm having trouble finding an equation that could relate temperature and my other given variables..I think I need to get the Work done, thus I can calculate the power...anyone care to enlighten me as to how to go about it?
 
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Hint: How much work is required to lift a 10 kg mass 4 m? Since the engine must produce that much work every second, how much power must it supply?

(I assume there is more to this question. Other parts will make use of the operating temperatures.)
 
You assume correctly(But I can do the rest myself), often times when reading a physics question over and over you still miss little things such as "per second", that are the keys to helping you solve a problem...Thanks Doc

I even typed it...sheesh...
 
[tex]w = m g \Delta h[/tex]
You'll get this in joules, times that one second gives you watts.
 

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