The first law says that the change in internal energy is equal to the work done on the system (W) minus the work done by the system (Q). However, can $Q$ be any kind of work, such as mechanical work? For example, does a string heat up if it is attached to a block and a tension is applied to move the block? And what about the work done on the system by gravity when an object falls? Is there any internal energy change when that happens?
[Physics] First law of thermodynamics applied to mechanical and gravitational systems
energythermodynamicswork
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Now we do $-ve$ work on the system while stopping it so the work done by the system is $+ve$. Is this part right?
Not always; you've stumbled upon a subtle point in the definition of work in mechanics, which is rarely discussed. Generally, if body $A$ has done work $W$ on body B, it does not follow that the body $B$ has done work $-W$ on the body $A$. This is true only if the two material points under action of the mutual forces have moved with the same velocity.
Explanation:
Definition of rate of work: When body S acts with force $\mathbf F$ on body $\mathbf{B}$ that has velocity $\mathbf v_B$ (more accurately, it is the velocity of the mass point which experiences the force), the rate of work being done by S on $\mathbf{B}$ is defined to be
$$ \mathbf F\cdot\mathbf v_B. $$ Notice that the velocity is that of the receiver, not that of the body the force is due to. So if I scratch my desk while the scratched portion remains at rest, no work has been done on the desk.
Thus, the definition of work is based on:
- the force due to the giving body;
- the velocity of the receiving body (its mass point the force acts on).
What about the work done on the body $S$? This is, per definition,
$$ -\mathbf F\cdot\mathbf v_S. $$
The forces have the same magnitude and opposite signs (due to Newton's 3rd law), but there is no general relation between the two velocities of mass points where the forces are acting. If $\mathbf v_S$ is not equal to $\mathbf v_B$ during the whole process, it is possible the two works done on the bodies will not have the same magnitude.
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Let's look at some specific cases.
Case 1. If a massive block $\mathbf B$ is brought to rest by another moving body $S$ with no sliding friction occurring (if the mass points of bodies experiencing the mutual forces always move with the same velocity), the velocities $\mathbf v_B,\mathbf v_S$ are the same and the two works have the same magnitude and opposite sign.
This is the case, for example, when the block is stopped in its motion by a spring mounted on a wall, or a person stops it gradually by hand. The kinetic energy of the block $E_k$ decreases to zero and equal amount of energy is added through work to the total energy of the stopping body. No heat transfer and no change in temperature need to occur, if there is no sliding friction and no temperature differences beforehand.
Case 2. If the block is stopped by forces of sliding friction - say, due to the ground - the description in terms of work is different. The mass point where the force due to ground acts on the block $\mathbf B$ is part of the block and is moving. Therefore the ground is doing work on the block (and from the reference frame of the ground, this work is negative). However, since the ground is not moving at all, the block does zero work on the ground!
This may look like violation of energy conservation, because block is slowing down without ground receiving energy, for there is no work received.
But it is only violation of mechanical energy conservation, which is fine and occurs daily. Total energy may still remain conserved, because it includes also internal energy of the block and internal energy of the ground, which change during the process.
As the block slows down, its kinetic energy transforms into different form of energy: internal energy of both the ground and the block. This happens with greatest intensity in the two faces that are in mutual mechanical contact.
The faces get warmer and for the rest of the system, they act as heat reservoirs. The energy is transferred via heat both upwards into the block and downwards into the ground.
Further , isn't work done always equal to change in kinetic energy (by the work energy theorem) ?
Only if the only energy that changes is kinetic energy. Generally, the work-energy theorem includes other energies. Kinetic energy can change into potential energy (in gravity field, in a spring) or into internal energy (inside matter, may manifest as increase in temperature or other change of thermodynamic state).
The answer really depends on the system under observation. The case given in your question in which change in internal energy is zero even when some work was done by (or on) the system is certainly possible if the system is not thermally isolated (or simple isolated). A system must not be thermally isolated because some heat energy has to transfer between the surroundings and the system, if we want internal energy to remain constant after some change in volume of system is observed.
From the first law of thermodynamics,
$\Delta U=Q-W_{system}$ $\tag 1$
If $\Delta U=0$ then,
$W_{system}=Q$ $\tag 2$
Equation $(2)$ implies that if in a closed (not isolated) system, the system expands, some heat comes into the system from the surrounding to replenish the internal energy lost when the system did some work against external pressure.
From the same equation, it also follows that if due to some external agent the system gets compressed, some heat gets out of the system to relieve the system of the internal energy it gained when the external agent did some work on the system.
In simpler terms, internal energy of the system increases when work is done on the system or heat comes into the system, and decreases when work is done by the system or heat gets out of the system.
If the internal energy has to remain constant, these two factors must work oppositely. Either one should increase the internal energy while the other decreases it.
In open systems, there is no boundary between the system and surrounding. Matter becomes exchangeable. In this case there will be no boundary for the system to perform the work against. The surrounding becomes the system. For open system the terms, $Q$ and $W$, have no significance.
Best Answer
The work in the first law is exactly the usual work $W=\int Fdx\rightarrow\int PdV$. For point particles, this is enough to completely specify the behavior of the system using Newton's first law, or energy methods. However, for macroscopic objects, the motion of the internal components (in thermodynamics these would be particles) have some additional degrees of freedom. Statistically, we can use the temperature to tell us things about "how much" energy the system has on its own, so changes in temperature can tell us how much its energy changes.
So the first law $\Delta U=Q-W$ tells us that we have to take care of heat transfer ($Q$) as well as the work ($W$) that is done on the system. Of course, these two quantities are not completely decoupled as I have described, but this is my intuition for how this works.