# Question on unit conversion with temperatures

1. Apr 20, 2013

### sgstudent

Hi, I was given 1Btu/(h.ft^2.°F) to W/m^2.K with these conversion ratios 1m^3=1000L 1Btu=1.055*103 1m=3.2808ft and Δ1K=Δ1.8°F

I can convert most of the units but I'm stumped by the temperature conversion here. Because I'm not converting a change in temperature but the discreet value of temperature here. So I'm not sure why I can just use the change in K to °F for that conversion. Shouldn't we use the conversion ratio for the discreet K and °F instead?

Hoping you guys can help me out with this :)

2. Apr 20, 2013

### jbriggs444

The quantities measured by the units you show, what do they represent?

I suspect that they are a measure of a surface's resistance to heat flow. You get a flow of energy that is proportional to time, proportional to surface area and proportional to the difference in temperature from the hot side to the cold side.

3. Apr 20, 2013

### sgstudent

Sorry I'm not sure what that equation represents. I was only asked in the question to change the units. But I'm not sure why I can use the Δ1K=Δ1.8°F for that though. Do you have any idea why I can do that?

4. Apr 20, 2013

### jbriggs444

The decision on whether to use delta temperature or absolute temperature rests on what the equation represents. In this case the relevant quantity is a temperature difference.

5. Apr 20, 2013

### sgstudent

Oh but I was told mathematically it is correct to use the change in K to °F even if the units given is a discreet K or °F. Is that wrong?

6. Apr 20, 2013

### Redbelly98

Staff Emeritus
To repeat what jbriggs said, these units refer to temperature differences, not actual temperature values. "1Btu/(h·ft2·°F)" means that the density of heat flow is 1 BTU/h per square meter of material, per °F of temperature difference between two sides of that material (through which heat is flowing).

It is simply not customary to include the "Δ" when expressing units like this. So we don't write it as BTU/(h·ft2·Δ°F). It's just what the convention is, and it is up to you to understand whether it is referring to a temperature difference (or change), or an actual temperature value.

Hope that helps.

7. Apr 20, 2013

### sgstudent

Hi thanks so much. This helps immensely :)

So if we were to know that the K is actually the discreet value and not the temperature difference, then we would have to use the other conversion ratio to solve it? So actually if I weren't taught that formula I wouldn't be able to solve it?

Also, why does temperature work in a different way as Litres. For example if we have a change in 50L and we want to change it to , we can just change it directly to cm3 we can just use the unit conversion of 1L=1000cm3 but for temperature we cannot do that?

And thanks so much. This has been bugging me ever since my tutorial.

8. Apr 20, 2013

### Redbelly98

Staff Emeritus
For discrete temperatures, it is not simply a conversion ratio. There is a formula, e.g. subtracting 273 from the Kelvin value to get Celsius, and then apply the usual conversion from °C to °F. But surely you have been taught this?

Not sure if I understand your question. Yes -- if you were not taught how to convert between different temperature scales, you would be unable to do so until you learned how. But that is an obvious thing to say, so I am wondering if you really meant that or something else?

Because 0 L is the same as 0 cm3, so converting the quantity works by using a conversion factor, just as converting a change in the quantity works.

But, for example, 0 °C is not 0 °F. So converting between °C and °F is more complicated than doing a simple multiplication.

9. Apr 22, 2013

### sgstudent

Thanks so much for the detailed explanation. What I meant was if I had a formula like PV=nRT and so if I were to convert the units of R from K to °F I would have to use the conversion of the discreet temperature values rather than the change in variables?

10. Apr 22, 2013

### MrAnchovy

Temperatures measured in Farenheit cannot be used in the equation PV=nRT (what would this imply at 0°F?)

11. Apr 22, 2013

### jbriggs444

Let me add a little detail to this.

Suppose that you had PV=nRT where T was expressed as a reading using the Kelvin scale and R was expressed in units of [mumble] per degree Kelvin. You could change from the Kelvin scale to the Rankine scale (the Rankine scale is Fahrenheit degrees above absolute zero). Then the units on R would change from degrees Kelvin to degrees Fahrenheit.

But if you wanted to use the Fahrenheit scale, that's not just a "linear" transformation. That's an "affine" transformation. The units on R would be the same, but you would need to change the form of the equation. In painful detail, that's...

PV = nRkelvinTKelvin
becomes... PV = nRRankineTRankine
becomes... PV = nRRankine(TFahrenheit+459.67)
becomes... PV = nRFahrenheit(TFahrenheit+459.67)

12. Apr 22, 2013

### sgstudent

Hi I'm not too sure what that all means. Why must I convert them to rankine first then Fahrenheit?

The question for the Units conversion for R was this 82.06(atm.cm3)/gmole.K to (psia.ft3)/lbmole.°F

So what I used was the conversion ratio of Δ1K=Δ1.8°F for this question. But I don't think I can do that because those temperature units doesn't represent the change in temperature right? So must I just substitute 1K for -457.87°F instead?

But the answer for that conversion for the R in the gas law was 10.73(psia.ft3)/gmole.K

Sorry for not understanding the explanation and thanks for helping me out.

Last edited: Apr 22, 2013
13. Apr 23, 2013

### MrAnchovy

This is incorrect, and is at the root of your confusion.

A BTU is the amount of energy required to raise the temperature of a cubic foot of water by one degree Fahrenheit, so it certainly does relate to a change in temperature.

The gas equation relates to absolute temperature and so can only be used with a scale that is zero at absolute zero. If you want to express the gas constant in a non-zero based scale such as Fahrenheit you need to apply the absolute zero correction to T to get PV=nRarbitrary(T-T0).

14. Apr 23, 2013

### sgstudent

Oh! But if i use the formula PV=nRΔT won't it be wrong? Because V is directly proportional to temperature and not change in temperature right?

Or if we set the initial temperature to be 0 degrees then we can use it?

But actually if something is non zero why can't we use it?

Last edited: Apr 23, 2013
15. Apr 23, 2013

### jbriggs444

Right. Because that formula expects temperatures to be expressed on an absolute scale.

For purposes of assigning units of measurement to conversion factors, that does not matter. The conversion factor for changing from one absolute scale to another is the same as the conversion factor for changing from one temperature difference measure to another.

And that makes sense. After all, an absolute temperature scale just expresses the difference between the measured temperature and absolute zero.

16. Apr 23, 2013

### MrAnchovy

There is no "initial temperature" in the equation PV=nRT: perhaps I confused you by using the symbol T0 to mean the value of absolute zero.

In the 18th century the work of Boyle, Charles and others explored the relationship between changes in pressure, temperature and volume of a gas and worked in Farenheit or Celcius/Centigrade; they observed the proportionality of changes but did not have an adequate model of the underlying physical processes.

In the 19th century the work of Joule and Thomson (who later adopted the title Lord Kelvin) revealed a new understanding of thermodynamics implying more fundamental scales of measurement which recognised the concept of absolute zero and the covertability of heat and mechanical work. It is no accident that we still use their names today when measuring these quantities in science, rather than units such as Farenheit and Horse Power.

17. Apr 26, 2013

### sgstudent

Hi sorry for bringing this up again, but when my teacher explained this question
(d) 82.06 (atm∙cm3)/(gmole∙K) to (psia∙ft 3)/(lbmole∙°F)
Given:
1m3 = 1000 L
1 kg = 2.2046 lb
1 in = 2.54 cm
1 m = 3.2808 ft
1 W = 1 J/s
1 Btu = 1.055 x 103 J
Change in 1 K = Change in 1.8°F
1 atm = 14.7 psia

To me, he explained that the fact that its a per kelvin as indicated by the divide line, it would instantly mean that the temperature units conversion are of "change in".

However, is this true? Because as MrAnchovy and the rest of you explained I cannot use Fahrenheit or degrees Celsius in the absolute form as that would mean that volume is is not proportional to temperature and the other laws where temperature is a variable.

So I was thinking that the better explanation was what MrAnchovy pointed one that in that question the temperature refers to T-T0. So in this case, it would make sense to use the change in conversion ratio.

Do you guys have the same idea here?

18. Apr 26, 2013

### MrAnchovy

This is ridiculous.
1. This is nothing to do with Mathematics
2. This is homework and should be in the appropriate section
3. The question you have been set is pointless - nobody uses units like those any more. There is some sense in working with conversions of things like BTUs, lb.ft, atm etc. as these are used for historical reasons in some branches of engineering. But lb-mole?
4. You simply cannot express the gas constant in Farenheit because the product of temperature and volume is not proportional to the temperature in Farenheit. If you want to express it on this scale you MUST use the Rankine scale. Whoever set this question is a fool.
5. 'He explained that the fact that its a per kelvin as indicated by the divide line, it would instantly mean that the temperature units conversion are of "change in"'. This explanation is rubbish. Does the "divide line" mean that there is a "change in" the number of molecules too?

19. Apr 26, 2013

### jbriggs444

The T in PV=nRT does not denote a change in temperature. It denotes a reading on an absolute temperature scale.

Yes. Since absolute scale readings are differences between temperature T and temperature T0 (aka absolute zero) the conversion factor to change from one absolute scale reading to another absolute scale reading is the same as the conversion factor to change from one temperature difference to another temperature difference. The conversion factor of 1.8 does work.

As MrAnchovy points out, you'd better be converting between Kelvin and Rankine, not between Kelvin and Fahrenheit though. Use of the Fahrenheit scale would be wrong, wrong, wrong. It's not an absolute scale.

Last edited: Apr 26, 2013
20. Apr 26, 2013

### sgstudent

Oops sorry for 1. Somehow I associated units conversion with mathematics. But also, thanks for the help! Your explanation is very clear.

Oh I think I get it now. The formula is an absolute temperature T so it wouldn't be totally correct to use T-T0 if I convert it to a Fahrenheit scale?

And I apologize for posting in the wrong sub-forums. Thanks for the help :)