Change in Temperature Equation HELP

In summary, the equation that the student was trying to solve was incorrect. They substituted all of the values correctly, but the answer came out to be too low. The missing zero was in the second term in the last line.
  • #1
FFX
8
0
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! :(


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  • #2
You should post this in the Homework and Coursework Questions subforum (read the forum rules).
 
  • #3
"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.
 
  • #4
dalcde said:
You should post this in the Homework and Coursework Questions subforum (read the forum rules).


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


Simon Bridge said:
"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.


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.
 
  • #5
Q/rcN=[itex]\frac{Q}{rcN}[/itex] or Q/rcN=[itex]\frac{Q}{r}cN[/itex]?
 
  • #6
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:
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.
 
  • #7
Thank you Simon!
I really appreciate the help! :)
 

1. What is the equation for calculating change in temperature?

The equation for calculating change in temperature is: ΔT = Tf - Ti, where ΔT represents the change in temperature, Tf represents the final temperature, and Ti represents the initial temperature.

2. How do you determine the units for change in temperature?

The units for change in temperature depend on the units used for the initial and final temperatures. For example, if the initial temperature is in degrees Celsius and the final temperature is in Kelvin, the units for change in temperature would be Kelvin (K). It is important to use consistent units throughout the equation.

3. Can the change in temperature equation be used for both increases and decreases in temperature?

Yes, the change in temperature equation can be used for both increases and decreases in temperature. The sign of the change in temperature (positive or negative) will indicate whether the temperature has increased or decreased.

4. Is the change in temperature equation applicable to all substances?

Yes, the change in temperature equation can be used for all substances as long as the initial and final temperatures are measured in the same units and the temperature change is within a reasonable range. However, for certain substances, such as gases, additional equations may be needed to account for changes in pressure and volume.

5. Can the change in temperature equation be used for non-thermodynamic systems?

No, the change in temperature equation is specific to thermodynamic systems where temperature is a key factor. It cannot be applied to non-thermodynamic systems, such as electronic circuits or chemical reactions, where temperature may not be the determining factor for change.

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