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Liquid in which vessel boils first?

  1. Nov 22, 2016 #1
    • Post moved from a technical forum, so homework template missing
    Hello ,

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    Above question came in a national level grade 10 exam recently .If suppose ΔQ amount of heat is transferred to B in time Δt , then how are we suppose to relate that to heat transferred to A . Should the rate of heat transferred to B be equal to that transferred to A ?

    I am attaching the official solution posted along with this question .

    It would be nice if experts could give their views.

    Many Thanks .
     

    Attached Files:

    Last edited: Nov 22, 2016
  2. jcsd
  3. Nov 22, 2016 #2

    phinds

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    That answer makes no sense to me. Does it make sense to you? Also, I think you and the "answer" are overthinking it. I don't see that it matters what either mass is. The only thing that I can see that matters is that the heat received by the water inside of vessel A has to first be absorbed by vessel B, then by the water in vessel B, then by vessel A, THEN by the water in vessel A. I can't see how that could possibly cause the water in A to boil first.
     
  4. Nov 22, 2016 #3
  5. Nov 22, 2016 #4

    russ_watters

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    Agreed.

    I'll say it another way: heat transfers from higher temperature to lower temperature areas. Therefore vessel B must always be warmer than vessel A.

    The answer assumes both that the mass of water in vessel B is larger than that in vessel A -- which doesn't have to be true -- and that the heat transfer rate is the same to both -- which most certainly is not true.
     
  6. Nov 22, 2016 #5

    jbriggs444

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    Perhaps I am missing something, but I see no diagram. It sounds like a classical double-boiler. The water inside the inner pan cannot boil unless and until all of the water in the outer pan has boiled away. Just as @russ_waters has stated. You use a double-boiler precisely because you do not want the inner contents to boil (or to caramelize on the inside of a pan).

    http://www.food.com/recipe/perfect-boiled-custard-20191
     
    Last edited: Nov 23, 2016
  7. Nov 22, 2016 #6

    haruspex

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    Me too. If vessel A has infinite conductivity then the temperatures inside and outside will be the same; anything less means a temperature difference has to exist to drive the heat flow.
    The offered solution looks like it belongs to a completely different problem: two vessels of known different mass contents with the same heat inflow.
     
  8. Nov 22, 2016 #7
    Thank you all for your precious views :smile:
     
  9. Nov 23, 2016 #8
    Well, late to the party as usual, but I will add my agreement. There is not the slightest justification for saying the heat flow into both containers is the same which is the entire premise of their solution. There are so many things wrong with this problem that is hard to talk about one of them because it sounds like you are implicitly accepting the others.

    A floats in B. Does that mean these are open vessels? Is there no heat lost out the top surfaces? Ok, they are closed vessels, they just have an air pocket. B is heated uniformly. The same heat rate over the whole surface? Do I then have to consider the gradient that will develop from where there is high heat capacity due to the water to where the heat capacity is low where there is only air? Ok, say they mean B is closed and uniformly in contact with a heat bath at an elevated temperature. Ahh boundary conditions on the whole exterior, now we can start reasoning.

    So the heat flow into B is equal to the heat flow into A. Wait!! What the hell?? A is inside B. If the heat flow into B is equal to the heat flow into A doesn't that mean that the heat is flowing straight through B and only into A??!?

    No, no! That would be silly. (but what else could they have been thinking when the set them equal). Anyhow, just look how they used the equations. A and B are heating up. Clearly they mean that the NET heat flow into B is equal to the net heat flow into A. So they are saying that 2Q/t are going into B and 1Q/t is going out of B and into A so that A and B have the same net heat flow. Wait, what? Why would that be true? What a coincidence!

    Hold on. They are using conductivity equations, so they are only talking about the boundaries. So the rate of heat flow across the outer boundary into B is the same as the rate of heat flow across the boundary into A. (And once again no heat stays in B for some reason) Well it's clear these should be equal because as you see they use the same delta T for both. Apparently the temperature drop from outside to B is inexplicably equal to the temperature drop across the boundary from B to A. Wait, what!!??

    Well hold on. It's perfectly fine that they randomly used the same delta T across both boundaries because they made up for it by icomprehensively using different times!! I am not making this up. Look at the solution. For some reason A and B have different clocks.

    So you see, the reason this was so confusing to all of us is that we didn't realize this is a problem in SPECIAL RELATIVITY!!! This might also explain why neither the water in vessel A nor the water in vessel B have any heat capacity. It has to do with time dialation and the twin paradox, but it's complicated so you wouldn't understand it.

    I'm sorry. This problem isn't right. It isn't even wrong.
     
  10. Nov 23, 2016 #9
    P.S.: my favorite thing about the solution is that A boils first. A's only contact with the universe is the water in B. It can never be hotter than B (in this case where B is being heated). What's more everybody knows this. The person who wrote the solution knows this. Every cook who has ever used a double boiler knows the water in A WILL NOT BOIL as long as there is still water in B in which to float.

    The temperature of A is at most the same as the water in B. The temperature of the water in B is at most the boiling point. If A is at the boiling point there is no net heat transfer into A. It can't boil.
     
  11. Nov 23, 2016 #10

    Nugatory

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    :smile: :smile:
    It seems a shame to propose an alternative to such an extravagantly excellent hypothesis... but I think that @haruspex called it right. The publisher messed up and the supplied answer belongs to a different problem.
     
  12. Nov 23, 2016 #11
    Yes, two different times to reach delta T sounds like two different trials with two different configurations. It must be a different question
     
  13. Nov 24, 2016 #12

    NascentOxygen

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    question-png.109298.png

    It is an ill-defined problem, I agree.

    But I think I could get the water in B to boil first, ahead of that in the outer vessel. Make the outer vessel a stainless steel sphere that is ⅓ filled with water. The inner vessel B is a little over ½ the diameter of A, and contains just a small amount of water, allowing B to sit high inside A. An array of burners applies heat to the outside of A uniformly.

    I haven't decided on the best choice for B, perhaps a sphere will do. Each container has a hole in the top.

    The upper wall of A can reach a relatively high temperature and radiant heating will, in turn, raise the exposed surface of B well above the boiling point of water. I'm anticipating this can occur before the large volume of water in A begins to boil.
     
  14. Nov 24, 2016 #13
    Relative to the posted problem and previous discussion swap all labels A and B in your description. I will swap them in my response below.

    I see what you are suggesting. I suspect you are right that in the extreme situation you could get some heat into the upper walls of A, the inner vessel . However that heat has to conduct down to the water, and whether a lot or a little all of the water in A is below the waterline on the outside of the wall. So doesn't the outside water boil too? Since it is higher on the heat gradient from the top of the wall doesn't it boil first?
     
  15. Nov 24, 2016 #14

    NascentOxygen

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    I'm planning for the greater mass of the outside water to delay its boiling. Stainless steel rather than copper because SS can support the heat gradient needed here. Experimentation may be needed, in particular with wall thicknesses and whether they should even be uniform.

    I'll avoid the A ↹ B nomenclature, then. :)
     
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