1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to use the first law of thermodynamics for simple mechanical problems

  1. Jan 19, 2015 #1
    I'm confused about what exactly is Q and U and their signs.

    Consider a block initially having some kinetic energy which we stop and we want to find by how much amount its temperature increases.

    Then Q=dU+W where dU is change in internal energy. (I'm using the sign convention that heat absorbed by the system is positive and work done by the system is positive).

    Now we do −ve work on the system while stopping it so the work done by the system is +ve. Is this part right?

    Now I'm really confused as to what to do next.

    I assumed system is not absorbing or giving out any heat so that dU=−W and since work is positive that means dU is negative and so this means temperature should very counter intuitively decrease.

    Isn't dU for the block mc(T(f)−T(i)) where T(f) is final temperature and T(i) the initial temperature , m is mass and c is specific heat capacity or is that wrong?

    A different approach I thought of was that dU=0 so Q=W and now we can let Q=mc(T(f)−T(i)) to find final temperature which'll come more than initial. This seems to work fine but I don't know why. Why should dU be 0 when temperature is increasing?

    The first approach seems to work fine with a different example of a human climbing an incline. There, work done by the human is positive and heat is 0 (or negative in the case human sweats, again i'm confused how should i go about deciding Q) So dU is negative (and even more negative if you consider sweating , again as expected) which is true since fat is used up.

    I'll provide one final example. Suppose a gas is in a container initially having some velocity. Then we stop it. We've to find the final temperature.

    Here again W is positive , now i'm again confused what to assume about Q. If we assume no heat exchange , Q=0 and so we've du=−W so dU is a negative quantity. But dU of ideal gas is 3/2nR(T(f)−T(i)) so this suggests final temperature is actually less and this seems wrong.

    I looked in different textbooks , but none has good discussion on this. Any help is appreciated.
     
    Last edited: Jan 19, 2015
  2. jcsd
  3. Jan 19, 2015 #2
    The equation you wrote for the first law is the form most frequently encountered and applied: ΔU=Q-W. However, this is not the form that applies in situations where there is a significant change in the kinetic energy and/or the potential energy of the system. In such situations, the more general form of the first law needs to be employed: ΔU+ΔKE+ΔPE=Q-W. This form of the first law is covered (usually briefly) in most textbooks on thermodynamics. For your problem, the equation reduces to:
    ΔKE = -W.

    And, ΔU=ΔPE=Q=0.

    Chet
     
  4. Jan 20, 2015 #3
    @Chestermiller Thank you for your answer. I'm having difficulty understanding why dU=0 . Consider that the block is stopped by friction. Then won't the block heat up when it stops and hence its internal energy as well as temperature should increase. In fact calculating the temperature increase is one of the problems in my textbook . How would one solve that using your form of the law?
     
  5. Jan 20, 2015 #4
    Further , isn't work done always equal to change in kinetic energy(by the work energy theorem) ? Or is that actually a subset of first law?
     
  6. Jan 20, 2015 #5
    In your original post, you didn't say that it was friction that was causing the block to stop. I assumed that you were exerting a force on the block to slow it down while doing work on another body.

    Doing a problem with friction at an interface (using the first law) is a very tricky. This is because the are discontinuities in the rate of heat flow and the rate of work being done on the two sides of the interface. I'll try to lead you through it.

    Let's first consider the side of the interface facing the table. If the table is insulated, the heat flow at this side of the interface (call it Q-) is zero. Since the table is rigid and not moving, the rate at which work is being done on this side of the interface by the table (call it W-) is also zero. So we have:

    Q- = 0
    W- = 0

    Let's next consider the side of the interface facing the block. Here the heat flow (call it Q+) is not zero since the heat generated is going into the block. In addition, the work on this side of the interface (W+) is not zero, since the friction force F is being applied to the block through a distance. The rate at which this side of the interface is doing work on the block is -Fd, where d is the displacement of the block, and, from a momentum balance on the block, this is also equal to ΔKE (of the block). So,
    W+= -Fd = ΔKE

    We can now apply the first law to the overall interface, including both the side facing the table and the side facing the block, to determine the heat flow Q+ from the block to the block-side of the interface. In carrying out this first law balance for the overall interface, we recognize that the interface has no mass, and so its change in internal energy and kinetic energy are zero. Therefore, we have:
    $$Q^+-W^++Q^--W^-=0$$
    So,
    $$Q^+=W^+=ΔKE$$
    Note that the heat flow and work in this equation are the heat flow from the block to the interface and the work done by the interface on the block. These are equal in magnitude and opposite in sign to the heat flow from the interface to the block Q and the work done by the block on the interface W. So, for the block, we have:

    Q = W = -ΔKE

    The first law for the block is ΔU + ΔKE = Q - W. Substituting the above values into this equation, we obtain finally:
    $$ΔU = -ΔKE$$

    I realize that this is very confusing and difficult to keep straight. It is important to keep close track of what is doing work on what, and what is transferring heat to what. I hope what I've shown makes sense to you. As I said, it's very tricky.

    Chet
     
    Last edited: Jan 20, 2015
  7. Jan 20, 2015 #6

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Chet,

    Thanks for the analysis. I agree with most of it but wouldn't the changes in potential, kinetic and internal energy of the body connected to the other side of the interface not have to be taken into account?I realize that you have said it is insulated, but that just means that the temperature on the surface will be very high. I don't think it means that no energy is imparted to that surface. If the body (the table + whatever it is connected to) is very massive there will be very little KE or PE imparted but I think that has to be addressed.

    AM
     
    Last edited: Jan 20, 2015
  8. Jan 20, 2015 #7
    @Chestermiller That was a very beautiful answer , thanks a lot! I've 3 small doubts.

    First , just to clarify your first statement , you mean that if instead of friction i stop the block slowly using my hand (hand is thermally insulated) then the block won't get heated up? If that is so , is that because the interface between hand and block does equal and opposite work on hand and block and since no heat goes to hand , no heat must go to the block because of the first law?

    Second , so the crux of the difference between friction and hand-force seems that friction does not do work on surface that is ground (earth ) assumed to be stationary. But since earth is so massive one can say that friction actually does do work on surface increasing its velocity by an immeasurably small amount and hence block gets no heat. What is wrong with that?

    Third , why is the work done equal to change in kinetic energy ? (I don't get what you mean by momentum balance ) Are you using work energy theorem here ? Or is that a subset of first law of thermodynamics?

    I'm very sorry for asking you this many questions and thanks a lot.
     
  9. Jan 20, 2015 #8
    Thanks Andrew. Good questions.

    First, with regard to the word "insulated." In the context that I used insulated, I thought it would be understood that this meant perfectly insulated (i.e., adiabatic). I believe that this was the spirit that the original homework problem was asked. So, no heat goes into the table. Of course, in real life, you are correct that the frictional heat generated will be parsed between the block and the table, in proportions depending on their thermal properties (thermal conductivity, heat capacity, and density). But, in this problem, it all the heat goes into the block.

    With regard to the work done on the table (and earth), this is just equal to the frictional force times the displacement of the table that occurs during the time that the force is non-zero. Once the frictional force drops to zero, no more work is done. This will translate into KE imparted to the table (and earth). But, since the displacement is so small during the time that the frictional force is non-zero, the KE imparted to the table will be insignificant compared to the original kinetic energy of the block.

    Chet
     
  10. Jan 20, 2015 #9
    I prefer to think of this more in terms of the block doing work on an ideal massless spring that gets compressed (and we lock the spring compression in place once the block has reached zero velocity). In this case, the kinetic energy of the block gets fully converted into compressional potential energy of the spring. So, W = -ΔKE, and ΔU=0.

    See my answer to Andrew Mason in the previous post.

    If we do a force balance on the block, we have:
    $$-F=m\frac{dv}{dt}=m\frac{dv}{dx}\frac{dx}{dt}=mv\frac{dv}{dx}=\frac{d}{dx}\left(m\frac{v^2}{2}\right)=\frac{d(KE)}{dx}$$
    So,
    $$W^+=-\int{Fdx}=Δ(KE)$$
    Note that this just follows from integrating Newton's second law (I think that this is the work energy theorem). It's often very helpful to apply what we learned in Mechanics to determine the work in thermo problems.
    No need for apology. This is what Physics Forums is all about.

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to use the first law of thermodynamics for simple mechanical problems
Loading...