Thermodynamics when work being negative or positive

Click For Summary
Work can be classified as negative or positive based on whether it is done by or on a system. When a system does work on its surroundings, it is considered negative, indicating energy loss, while work done on the system is positive, indicating energy gain. The first law of thermodynamics is applied using the equation ΔU = Q - W, where a positive work value signifies energy expended by the system. In the given example, 400 J of work done by the system results in a decrease in its internal energy. Understanding these sign conventions is crucial for accurately analyzing energy changes in thermodynamic processes.
manal950
Messages
177
Reaction score
0
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
 
Physics news on Phys.org
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 19 ·
Replies
19
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
2K
Replies
7
Views
2K