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Change in Temperature Equation HELP!

  1. Jan 22, 2012 #1


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    Hey Everyone,

    First time posting here, but you'll probably see me around quite a bit!
    If this is in the wrong math spot please let me know, I don't understand the different types of maths.

    I have an equation on the change of temperature, I substituted all the values in correctly and thought I was working it out correctly, but my answer came out far too low for me to think I did it correctly. If someone could have a look and give me guidance, that'd be great. The first picture is the full question, the second is me 'trying' to work it out! :(


  2. jcsd
  3. Jan 22, 2012 #2
    You should post this in the Homework and Coursework Questions subforum (read the forum rules).
  4. Jan 22, 2012 #3

    Simon Bridge

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    "Put this into MS Excel"
    .... evil! I take it this is a course on how to use Excel?

    Anyway - need to know how you went about it and how you know it came out too low.
    That formula is utterly shocking - we'd normally simplify it.

    looks like you didn't enter values correctly - I can see from 2nd to last line to last line you've misplaced a zero for example. But it is hard to tell.

    You should use separate cells for your input data, then use the cell names as variables in your formula.
  5. Jan 22, 2012 #4


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    Oh I didn't notice that sticky, I'll post it in the homework area.

    It's not a course on how to use excel, the first part (part A) is supposed to be done just manually. The second part is supposed to be done on excel I presume. I don't even know how to use excel. I'm doing this math unit externally so it makes it a lot harder as I don't have a teacher telling me how to do anything.

    I can't see the misplaced 0, could you be more specific on where it is please? Sorry.

    I was just working it out following the basic BODMAS thing, brackets first etc.

    I don't have the final answer there because already you can see it's going to come out to a very small number. I presume the question would be absolutely retarded if the temperature was only like 0.0000001 of a change lol.
  6. Jan 22, 2012 #5
    Q/rcN=[itex]\frac{Q}{rcN}[/itex] or Q/rcN=[itex]\frac{Q}{r}cN[/itex]?
  7. Jan 22, 2012 #6

    Simon Bridge

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    OK: so where it says "use program xyz" just use the program you have.
    You could use OpenOffice-calc instead or any spreadsheet .... or matlab, or octave.

    It's all multiplication and division so BODMAS may not get you into to much trouble ... bear in mind that a common error is to put 5-3+2=0 since it says to do the addition first. We can get rid of BEDMAS by being more careful how we write relations so 5-3+2 = (+5)+(-3)+(+2) and instead of divided-by signs we write fractions and so on.

    Conventionally a/bc would be the same as a/(bc) ... the equation writer has been careful with 1/(2(Qs/Q)) and (r/c)f after all. (OTOH: has not been careful with variables - there is a r in the equation and an R in the text... guessing r=R?)

    The missing zero, I think, is in the second term in the last line.
    Code (Text):
    octave:174> 440.022/(23.43/0.00000002556)
    ans =  4.8002e-07
    ... you have 4.8002e-06

    I think you can expect a very small change in temperature from those figures and that equation - you have to do your reality-check against the physics: what is the situation that leads you to expect a larger temperature change?

    If I put k=8.3e-06, and write Qs as Qs (for clarity) then that equation can be rewritten:[tex]\Delta T = \frac{kr^2Nf}{Q(1-\frac{Q}{2Q_s})}=\frac{2Q_skr^2Nf}{Q(2Q_s-Q)}[/tex]... (check my algebra) you may find that easier to handle.
    Note: the dimensions don't pan out - so either some units are missing ([k]=m-2 perhaps?) or the equation is wrong.
  8. Jan 23, 2012 #7


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    Thank you Simon!
    I really appreciate the help! :)
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