some math twisters?

[wolfson]

Yes as i already stated there are two possible answers the Physics one and the mathematical one, I chose to represent the mathematical version you chose the physics!!!!!!!!

[YT2095]

twice as cold compared to WHAT tho?

[Sayonara]

The explanation "it's the mathematical answer" makes no sense.

 

YT: Who are you talking to?

[YT2095]

I was addressing you sayo

 

my post fell kinda out of sequence is all

 

hehehehe Who would have thought we`de have been above the 50`s in post with this peice of junk tho! )

 

[Sayonara]

* bangs head on wall *

 

In the only interpretation of the question that makes sense, "twice as cold as 0c" is considered to be the temperature when the energy of the system falls to half its original level.

 

This is best described in Kelvin. It can be described in celsius or Farenheit, but that does not mean that we randomly base the calculation on "zero". Hence the bit where I said "poorly-applied physics".

 

 

If one uses any other interpretation of the question, the premise will always be flawed. Therefore whatever one does one will be "wrong", even if correctly applying the maths and/or physics.

[atinymonkey]

I think your both right, just one of you is half as right as the other. Or is it one of you is twice as right as the other?

 

Maybe it's both.

[wolfson]

Ok say^3 why dont you ask Maths Professor, and a physics Professor and then identify the answers, there is not ONE correct answer!!!!!!!

[atinymonkey]

Well, your just all wrong. Ask Mr Hoppy.

[Sayonara]

wolfson said in post #57 :

Ok say^3 why dont you ask Maths Professor, and a physics Professor and then identify the answers, there is not ONE correct answer!!!!!!!

You can make up any random calculation you want and correctly solve it, but that doesn't necessarily constitute a "right answer". Half the problem is working out the question.

[YT2095]

there maybe 3 correct even if the next day is still at 0c but twice as long!

[Sayonara]

That's an entertaining thought, but not a likely scenario.

[YT2095]

esp if you do Shift Work

[VendingMenace]

WOW, what a debate. I will try to explain this problem some.

 

Yes as i already stated there are two possible answers the Physics one and the mathematical one

 

Ok, i will play your game

 

so you claim that there are two answers, the physical one and the mathmatical one. For the physics one, i think we can all agree that you must convert to kelvin first. (at least i hope this should be clear). This is becuase temperature is a measure of average kenetic energy of the molecules in the system. Half the temp means half the energy. Period. The easiest way to approach this is to convert temp to kelvin and divide by two. This is how the problem must be solved physically speaking.

 

I chose to represent the mathematical version

 

Ok then, lets look at a mathmatical answer. In a purely mathmatical system, physical ideas hold no significance, we are working purely in the realm of numbers. As such, celcius, farenhieht and kelvin hold no meaning -- we must ignore them. Becuase of this, we cannot convert between any two temp scales (if we tried, we would be admiting that the numbers have physical significance, wich they don't in a purely mathmatical appraoch).

 

Thus, we are ask what is half of 0. The only answer can be 0/2 or 0. Thus, -8.9C is incorrect, as that is an answer that has physical significance.

 

The only purely mathmatical answer is 0. The only phsically correct answer is -136.575C.

 

I hope that makes sense.

[atinymonkey]

The application of mathematics does not preclude the world of physics. As the measurement of temperatures is a mathematical principle, the idea that a mathematician would not use conversion to produce the solution is a simplistic view of the world.

This is all just a pedantic argument to answer what was not really a conundrum, but a basic test of the ability to convert measurements of temperature.

[Sayonara]

VendingMenace said in post #63 :

so you claim that there are two answers...

Ok then, lets look at a mathmatical answer. In a purely mathmatical system, physical ideas hold no significance, we are working purely in the realm of numbers. As such, celcius, farenhieht and kelvin hold no meaning -- we must ignore them. Becuase of this, we cannot convert between any two temp scales (if we tried, we would be admiting that the numbers have physical significance, wich they don't in a purely mathmatical appraoch).

See, my point here is that - based on the scant information in the teaser - randomly deciding to apply pure maths and trying to divide zero by 2 is not answering the question. Or at least, not answering any interpretation of the teaser that makes sense. You've provided "an answer", sure, but not one that's any help.

 

It's a bit like me asking someone at a bus stop how frequently the buses run, and instead of them working it out based on whatever evidence is at hand they multiply the current time by 10 and claim that's the answer. Or like atinymonkey standing just over there --> shouting "the answer is green!" because that's the colour of the bus.

 

Like MrL says, there is no mathematical answer because "cold" is not a mathematical function.

[VendingMenace]

The application of mathematics does not preclude the world of physics.

 

I am not claiming that it does, i am merely trying to show that to get an answer other than -136C you must ignore the physical reality of the system. Once you do that, the only thing left to let you solve the problem is mathmatics. Thus, and answer gartered this way will be a purely mathmatical answer (ie. no physical significance).

 

As the measurement of temperatures is a mathematical principle

 

No, it is a physical priciple that happens to use mathmatics. Energy is not a mathmatical priciple any more than mass is. The things we measure are physical properties wich we express using mathmatics.

 

the idea that a mathematician would not use conversion to produce the solution is a simplistic view of the world.

 

I am not claiming that a mathematitian would ignore the physicallity of the problem. Again, i am merely trying to show how to logically arive at an answer other than -135 C and the only way is to ignore the physics of the problem.

 

Of course, as sayonara pointed out, this doesn't really answer the question, but more on this later....ok, now.

 

there is no mathematical answer because "cold" is not a mathematical function.

 

perhaps i was unclear, i am sorry. This is the point i was trying to make. They only way you could every come up with an answer other than -136C is to ignore the physical significance of the question. Then you may get another answer.

 

Of course, this answer has no physical meaning, that is why i said the answer would be 0. Not 0 degrees C. If you try to answer the question with just pure math, then you loose the physcis of it and left with an answer that does not really answer the question :/

 

Thanks for making that really clear though SWEET!

[Sayonara]

But just saying "zero" doesn't answer anything.

[YT2095]

ok, rather than tackle this arg from many different fronts, lets address a single arg at a time

 

what is wrong with converting to F from C performing a calc then converting back?

[wolfson]

Nothing at all YT

[Sayonara]

It doesn't answer any question that (a) can be derived from the teaser and (b) makes sense.

[wolfson]

I give up

[VendingMenace]

what is wrong with converting to F from C performing a calc then converting back?

 

Physically this is incorrect. I will try my best to give a detailed explination of why, it will out of nessesity use a fair amount of math. However, it will only be basic algebra, so while it might be somewhat tedious, please follow along if you really want to understand.

 

 

First off there are some things that we must realize;

 

First, Temperature is a measure of average kenetic energy of the molecules in teh substance you are measuring the temperature of. The eqation that relate kenetic energy to temperature is as follows;

 

KE=(3/2)kT

 

Where:

KE=kenetic energy

k=boltzmann’s constant

T= temperature (in Kelvin)

 

You may wonder why the temperature has to be in kelvins for this equation. Wait just a bit and you will understand. (the answer to this is why you cannot just convert to F from C, perform the cal then convert back).

 

 

Second, The kentic energy of an object is related to the speed at wich it is traveling. The eqation that relates energy to speed is as follows...

 

KE=(1/2)(mv²)

 

Where:

KE=kenetic energy

m = mass of particle

v = velocity (but since the velocity is squared, then the only important part is the speed and not the direction.)

 

looking at this equation, we find that if we have two particles of a given mass and one is traveling twice as fast as the other, then the faster travelling particel will have 4 times as much energy. Correct?

 

 

 

Ok, now that we have got the preliminaries out of the way, we can get down to business.

 

First off, we can combine these two equations for kenetic energy together. If we realize that we must arrive at the same value for the energy whether we calculate the kenetic energy of sample by mearuring its temperature or figuring out the averate speed of the molecules in it, then we can set the two equations equal to eachother. That is;

 

(3/2)kT= KE =(1/2)(mv²)

 

(3/2)kT=(1/2)(mv²)

 

using this new equation, we can solve for how the speed of molecules vary if we were to change the temperature. We find that

 

(3/2)kT= (1/2)(mv²)

 

(2)(3/2)kT=(mv²)

 

(3)(kT/m)=(v²)

 

[(3)(kT/m)]^1/2=v

 

 

so we find that the speed of particles in a substance is related to the temperature by the following way…

 

v=[(3)(kT/m)]^1/2

 

cool.

 

HERE IS THE ANSWER

Now we see why we must use the Kelvin scale to calculate energies. If we used either C or F we could have a negative temperature on the right side of this equation. The boltzman constant is a positive number as well as the mass. Thus, if the temperature were allowed to be negative, we would be trying to take the square root of a negative number! This would lead us to a an answer than had an imaginary number. But speed is not an imaginary thing, so we cannot ever have an imaginary value.

 

This is, in essence, why you must use Kelvin when calculating “how cold is half” and indeed, if you do not wish to make a mistake, you should always use Kelvin when dealing with questions that ask you to manipulate or use temperature.

 

 

Well, I really hope that clears up the questions you all have. It certainly was a lively debate. And feel free to ask more questions, questions are good!

[VendingMenace]

I was thinking that perhaps some people would not be satisfied by this mathematical and physical explination given above? So I decided to work through an example real quick, for those continueing doubters (I am not trying to be mean or antagonistic here. I think that it is good to insist on explinations until you feel that you have a good understanding of something).

 

 

Let’s see then, what question should we use. Oh yes, lets answer the following;

 

if the ouside temperature is 0c.

and it will be twice as cold tommorow.

What temperature will it be?

 

 

Lets assume that you go ahead and do it incorrectly, and convert from C to F, do your calc and then concert back to C again. Well we have already seen that the answer you will get then is, -8.9 C. Now let’s see if this answer holds up to the math, shall we?

 

Again we have the equation;

 

v=[(3)(kT/m)]^1/2

 

 

as proved above, we must use K in this equation, so lets convert -8.9C = 264.1 K (I just used 273 as the conversion, for simplicity)

 

pluging in the numbers (boltzmann’s constant, k=1.38 x 10^-23 J/K). The identity of the material doesn’t matter, right, so lets assume that it has a mass of 1x10^23 kg.

 

v=[3(1.38x10^-23 J/K)(264.1 K)/(1x10^-23 kg)]^1/2

v=33.1 m/s

 

Ok, so we have the velocity of the sample at -8.9 C. WE using the same exact process, we can calculate the sample’s temperature at 0 C, that is, its starting temperature. 0 C =273K. SO we find that the velocity of the molecules at 0 C would be…

 

v=33.6 m/s

 

 

 

now the original question was aking for half the temperature right? That means that nothing but the temperature changes. So we look at the equation;

 

v=[(3)(kT/m)]^1/2

 

and ask ourselves what will happen to the speed of the molecules when we drop the temperature by half? I think that it is fairly easily seen that the speed will decrease by a factor of (1/2)^1/2.

 

 

So the average velocity after cooling should be (1/2)^1/2 the original velocity at 0 C.

 

 

So, we see that we should have;

 

(1/2)^(1/2)(velocity at 0 C) = (velocity at -8.9 )

 

 

plugging in our values we obtained above, we have;

 

(1/2)^(1/2)(33.6 m/s) = (33.1 m/s)

 

23.75 m/s = 33.1 m/s

 

This is obviously not true. And so we see that we do not arrive at the correct answer by converting to F then carring out calc and converting back to C.

 

 

Whew! That was a lot of typing, and I think I am done for now. I hope that answers your questions. The take home message is that you must convert to K before carrying out the calculations. You will run into similare problems everytme you do not do this. This is partially why the Kelvin scale was invented. Basically, the Kelvin scale takes into account the baseline energy for you, so that you do not have to think about it. Fail to do this, and you run into serious problems.

 

Well that is it for now!

[jordan]

Nicely donce VM. Assuming all your calculations are correct, that made a lot of sense to me. I'm glad you took the time to explain all of your steps or else that would have all been over my head.

[Sayonara]

Fantastic response!

 

I'm amazed you took the time VM