Thermodynamics when work being negative or positive

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Homework Help Overview

The discussion revolves around understanding the sign conventions for work and heat in thermodynamics, particularly in the context of the first law of thermodynamics. Participants are exploring how to determine whether work is considered negative or positive based on the system's interactions with its surroundings.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the criteria for determining the signs of work and heat, with some suggesting that work done by a system is negative while work done on a system is positive. Others are examining the implications of these conventions on the internal energy of the system.

Discussion Status

There is an ongoing exploration of different interpretations of the sign conventions. Some participants have provided insights into how the first law of thermodynamics applies, but there is no explicit consensus on the correct application of the conventions. Guidance has been offered regarding the implications of positive and negative work on the system's energy.

Contextual Notes

Participants are navigating potential misunderstandings about the sign conventions and their impact on calculations, particularly in relation to the first law of thermodynamics. The original poster's question indicates a need for clarification on these conventions.

manal950
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Hi

How are you

I have a question how we can know the work if - or +

also heat energy

please could see this question why in answer
take work as -
and heat as +

857210025.jpg
 
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hi manal950! :smile:
manal950 said:
I have a question how we can know the work if - or +

it's really english rather than physics …

work done by a system is minus

work done on a system (or to a system) is plus :wink:

the question says "400 J of work is done by the system", so the work comes out of the system, so it's minus

(if the question said "400 J of work is done on the system", then the work would come into the system, so it would be plus)​
 
Tim got it backwards for the sign convention you're using. You have the first law written as ##\Delta U = Q-W##. Assume for a moment that Q=0. If work is done by the system, the system is expending energy on its surroundings, so the system's energy U must decrease. For this to happen, you need W>0. Similarly, if the surroundings do work on the system, energy is being added to the system, so you want W<0 so that ##\Delta U>0##.
 
thanks so much

thanks tiny-tim
thanks vela
 
If the work was negative, that is -400J, then the answer would have been

ΔU = Q - W
= 1000 J - (-400 J)
= 1000 + 400
= 1400 J

So negative work means that the system gained energy or an external force did work on the system.
In this case the problem states that the work is positive 400 J. Which means that the system did work
and therefore lost energy indicated by a lower change in its internal energy.
 

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