Understanding q: Why does dqsurr = -dqsys?

In summary: You would have to quote the source. Either it is wrong, it is about a specific situation or you have misunderstood it.
  • #1
santimirandarp
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The question is:

why ##dq_{surr}=-dq_{sys}##?

q=heat, surr=surroundings, sys=system.

Is there any simple way to understand this?
 
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  • #2
santimirandarp said:
The question is:

why $dq_{surr}=-dq_{sys}$?

q=heat, surr=surroundings, sys=system.

Is there any simple way to understand this?

Conservation of energy?
 
  • #3
PeroK said:
Conservation of energy?

How do you derive that expression from the conservation of energy?
 
  • #4
santimirandarp said:
How do you derive that expression from the conservation of energy?

If the heat (energy) leaves the system it must go to the surroundings; and, vice versa. The change in one must be equal and opposite to the change in the other.
 
  • #5
PeroK said:
If the heat (energy) leaves the system it must go to the surroundings; and, vice versa. The change in one must be equal and opposite to the change in the other.
I'm sorry but I don't understand that answer
 
  • #6
santimirandarp said:
I'm sorry but I don't understand that answer

Is your problem conceptual or mathematical?

Do you understand the concept of conservation? It means that the total amount of something stays the same over time.

Mathematically this means that if you add up all the changes in something it must come to zero. In this case:

Change in energy (system) + change in energy (surroundings) = 0

or:

##dq_{sys} + dq_{sur} = 0##
 
  • #7
PeroK said:
Is your problem conceptual or mathematical?

Do you understand the concept of conservation? It means that the total amount of something stays the same over time.

Mathematically this means that if you add up all the changes in something it must come to zero. In this case:

Change in energy (system) + change in energy (surroundings) = 0

or:

##dq_{sys} + dq_{sur} = 0##

I know maths and conservation.
But why do you say heat is conserved?
 
  • #8
santimirandarp said:
I know maths and conservation.
But why do you say heat is conserved?

Because that's what you said in the original post. Your equation is equivalent to conservation of heat energy.

Say, for example, you have:

Change in the number of apples in the system = - change in the number of apples in the surroundings

And you asked "is there a simple way to explain this?" Then I'd say "conservation of apples".
 
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  • #9
PeroK said:
Because that's what you said in the original post. Your equation is equivalent to conservation of heat energy.

Say, for example, you have:

Apples put in the barrel 1 = - apples put in barrel 2

And you asked "is there a simple way to explain this?" Then I'd say "conservation of apples".

It is not. The equality I showed is exactly what I don't understand ('why ...'). So I'm asking: where does the equality comes from? (and heat is not so simple as apples).
 
  • #10
santimirandarp said:
So I'm asking: where does the equality comes from?

It's the first law of thermodynamics.
 
  • #11
PeroK said:
It's the first law of thermodynamics.
It is not. The first law implies that universe internal energy is conserved and also:

##\Delta U_{sys}=(W+Q)_{sys}=-\Delta U_{surr}=-(W+Q)_{surr} ##

so ##(W+Q)_{sys}=-(W+Q)_{surr} ##
I don't see how it follows that dq=-dq_{surr}
 
  • #12
santimirandarp said:
why $dq_{surr}=-dq_{sys}$?
This is only true in the absence of any work.
 
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  • #13
DrClaude said:
This is only true in the absence of any work.

I've found it quoted as a general truth. Thanks.
 
  • #14
santimirandarp said:
I've found it quoted as a general truth. Thanks.

Are you telling us or asking us? That this is a universal truth? And, that the obvious explantion that it's the first law in the absence of work won't do?
 
  • #15
PeroK said:
Are you telling us or asking us? That this is a universal truth? And, that the obvious explantion that it's the first law in the absence of work won't do?

I'll ask the second part on a different question.
 
  • #16
santimirandarp said:
I've found it quoted as a general truth. Thanks.
You would have to quote the source. Either it is wrong, it is about a specific situation or you have misunderstood it.
Anything that’s a "general truth" will be stated in many places.
 
Last edited:

1. Why is dqsur = -dqsys in thermodynamics?

In thermodynamics, dqsur and dqsys represent the heat transfer between a system and its surroundings. The negative sign in the equation indicates that heat is flowing from the system to the surroundings, leading to a decrease in the system's internal energy.

2. What is the significance of dqsur = -dqsys in thermodynamics?

The equation dqsur = -dqsys is a fundamental principle in thermodynamics known as the Second Law of Thermodynamics. It states that heat naturally flows from hotter objects to colder objects, and this flow of heat is irreversible.

3. How is dqsur = -dqsys related to entropy?

Entropy is a measure of the disorder or randomness in a system. The equation dqsur = -dqsys can be used to derive the expression for entropy change, dS = dqrev/T, where dqrev is the reversible heat transfer and T is the temperature. This shows that the Second Law of Thermodynamics is closely related to the concept of entropy.

4. Can dqsur = -dqsys ever be positive?

According to the Second Law of Thermodynamics, the equation dqsur = -dqsys is always negative, indicating that heat flows from the system to the surroundings. However, in certain cases, such as during a phase change or in a heat pump, dqsur can be positive, but only if work is done on the system to transfer heat from the surroundings.

5. How does dqsur = -dqsys affect the efficiency of a heat engine?

The equation dqsur = -dqsys plays a crucial role in the efficiency of a heat engine. The efficiency of a heat engine is given by the ratio of work output to heat input, which is always less than 1. This is because some heat is always lost to the surroundings, as indicated by the negative sign in the equation. Therefore, the Second Law of Thermodynamics limits the efficiency of heat engines.

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