Thermodynamics when work being negative or positive

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SUMMARY

The discussion focuses on the sign convention for work in thermodynamics, specifically how to determine whether work is positive or negative. When a system does work on its surroundings, it is considered negative, while work done on the system is positive. This is illustrated through the first law of thermodynamics, represented as ΔU = Q - W. For example, if 400 J of work is done by the system, it results in a decrease in internal energy, while negative work indicates energy gain for the system.

PREREQUISITES
  • Understanding of the first law of thermodynamics (ΔU = Q - W)
  • Familiarity with concepts of work and energy in physics
  • Knowledge of sign conventions in thermodynamic processes
  • Basic grasp of heat transfer and internal energy
NEXT STEPS
  • Study the implications of the first law of thermodynamics in various scenarios
  • Explore examples of work done by and on a system in thermodynamic processes
  • Learn about different types of thermodynamic systems and their energy exchanges
  • Investigate the relationship between heat transfer and work in closed systems
USEFUL FOR

Students of physics, engineers working with thermodynamic systems, and anyone seeking to understand energy transfer and work in thermodynamics.

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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