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vovka

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  1. Hi! Really appreciate your wide thoughtful participation in this topic studiot. I study general physics ( thermodynamics in particular). Thank you for kind proposition about scans but I was looking for textbook in thermal physics (thermodynamics and kinetic theory of gases). I have a few (though truly demanding but think worth to give a try) and the above mentioned Cengel and Boles seems to be good additional reference.
  2. Many thanks for your careful review, studiot. I agree we can express heat through enthalpy \({δQ=dU+pdV=dH}\), and since the body is hardly compressible then variations in V and p are negligible so there is no work done. From (2.28) we have \({ dQ=dU=dH=CdT}\) that leads us to integration. We don't have to specify what kind of C we use since \({C≈C_V≈C_P}\). Finally from the above I conclude that the solution approach is mainly correct and the discrepancy should be referred to misprinting or lack of info in the problem. (taking \({C_V}\) implies V=const so pdV=0 and we shouldn't include any work done) p.s. could you name the book you've posted above
  3. Thanks for reply, studiot. The equation is as written in the textbook, \[{C=10+0.002T+3*10^{-5}T^{2} JK^{-1}}\] I think T plays role here as some independent variable in [300,400] and we should treat it as simple number but your argument is logical. So far I have discovered some misprints in this textbook thus it may be here as well.
  4. \(\sqrt{\pi})\ \(\sqrt{\pi}\)
  5. Hello! The problem is stated as follows: The heat capacity of a body, in the considered temperature interval, depends on temperature according to the expression C=10+0.002T+3*10^{-5}T^{2}JK^{-1} How much heat is released when the temperature varies from T1= 400 K to T2=300 K ? My way of solving: we have C=δQ/dT so δQ=CdT . By integrating both sides on the given interval we should get the desired quantity. My computation yields -1440J but the answer is -2.1kJ?! What's wrong? One remark is that δQ is not a differential so taking integral of it may be not correct. Will be grateful for reasonable opinions. Thanks for attention!
  6. \( y=x^2 \) x^{-2}
  7. The point here is 9a^2=3q^2 leads to 3a^2=q^2 so q=3b, b ∈Z thus p/q=3a/3b which contradicts with our assumption that p/q is irreducible.
  8. Hello guys! I've faced a minor difficulty in proving the statement "√3 is irrational number". My arguments are the following: Let's suppose that √3 is rational then it can be expressed as p/q, p,q∊Z which is irreducible, so √3=p/q <=> 3=p2/q2 <=> p2=3q2 And here is a problem from p2=3q2 I concluded that p=3a, a∊Z after that the proof is led to contradiction => p/q can be reduced by 3. The ground for my doubt is we cannot conclude from product's divisibility the divisibility of it's factors (maybe even these factors are the same) e.g. 12 | 3*8 but 12 ∤ 3 and 12 ∤ 8. I apologize for my English)) and appreciate any attention.
  9. Hi dude! I was concerned with the same questions and just surfed the internet...I picked up some books and started working through. The main question is what is the best choice for self studying, and the answer varies from person to person depends on his background and goals. Certainly you should have a calculus book (there are plenty of them in net), some general physics book (like 'fundamentals of physics' by Halliday and Resnick) and probably linear algebra book. If your knowledge in elementary algebra and in trigonometry is not strong enough you should look for some as well. If you want I can give you some pdf-copies I use or the reference to the source I got them... Good luck with your studying.
  10. vovka

    homework help

    I don't agree with answer. I think 'inverse' forgot to change sign...
  11. I use it, but some limits cannot be solved by this mean (it has a good graphing utility instead and you can figure out needed value). Also I use another one placed in onsolver.com and don't have any trouble to this day except for poor understanding about this kind of matter
  12. Hi everyone. Is there any good FREE on-line limit solver? Thanks. I've already found one....
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