Conversion of Temperature Scales

[whoareyou]

Can someone explain the last paragraph of this slide? It doesn't make sense to me.

 

[berkeman]

Can someone explain the last paragraph of this slide? It doesn't make sense to me.

First, since it is addressing rates, the constants are no longer needed (does that make sense?).

Second, each of those fractions is equal to 1. Just like 1km/1000m = 1. You can use fractions like these to convert from one unit to another, or one rate to another involving the units.

So if you have a rate like 0.2mV per degree Celcius, you could convert that into a rate involving the Rankine temperature scale if that were of some help for some reason. Which one of those fractions would you use for such a conversion?

 

[whoareyou]

First, since it is addressing rates, the constants are no longer needed (does that make sense?).

Second, each of those fractions is equal to 1. Just like 1km/1000m = 1. You can use fractions like these to convert from one unit to another, or one rate to another involving the units.

So if you have a rate like 0.2mV per degree Celcius, you could convert that into a rate involving the Rankine temperature scale if that were of some help for some reason. Which one of those fractions would you use for such a conversion?

I get how to do the converions, I just can't quite understand why when you want to convert to another temperature scale, you don't use the first set of equations listed. I mean 0°C != 1.8(32°F).

 

[berkeman]

I get how to do the converions, I just can't quite understand why when you want to convert to another temperature scale, you don't use the first set of equations listed. I mean 0°C != 1.8(32°F).

That's the rate issue that I mention in the first part of my post. If you differentiate an equation, what happens to the constants? The paragraph is discussing when you are dealing with rates that involve temperatures.

My example of 0.2mV per degree C is the temperature coefficient of voltage in a diode junction...

 

[whoareyou]

That's the rate issue that I mention in the first part of my post. If you differentiate an equation, what happens to the constants? The paragraph is discussing when you are dealing with rates that involve temperatures.

My example of 0.2mV per degree C is the temperature coefficient of voltage in a diode junction...

Constants become 0 when differentiated. But ... I don't get it lol :(

Btw, your example would be 0.2mV/°C x 1°C/1.8°R = 1mV/9°R.

 

[lurflurf]

So if we have a temperature difference we do not care about the values only the differences thus for differences
°R= °F
K= °C
Thus we can use ratios like for absolute temperature

 

[berkeman]

So if we have a temperature difference we do not care about the values only the differences thus for differences
\Delta °R= \Delta °F
\Delta °K= \Delta °C
Thus we can use ratios like for absolute temperature

I've added in delta symbols to show that we are talking about changes...

 

[whoareyou]

Ok I see how the math works out with temperature changes. But in this example:

It's not a temperature change is it (ie. ΔK to Δ°R)?

 

[lurflurf]

K and °R are absolute temperature scales so conversion can be done by ratios. Absolute temperature can always be regarded as a change, that is absolute temperature=Absolute temperature-temperature absolute zero. °F and °C are not absolute so conversion requires also addition and subtraction. Temperature conversion in general can be effected by two reference temperatures say A and B
$$T^\prime =T_A ^\prime + \frac{T_B ^\prime-T_A ^\prime}{T_B -T_A }(T-T_A)$$
primes denote measurement in the new system
example
$$T_{human} (\,^{\circ}\mathrm{F}) =T_{water freeze} (\,^{\circ}\mathrm{F}) + \frac{T_{water boil} \,^{\circ}\mathrm{F}-T_{water freeze}\,^{\circ}\mathrm{F} }{T_{water boil}\,^{\circ}\mathrm{C} -T_{water freeze} \,^{\circ}\mathrm{C}}(T_{human}\,^{\circ}\mathrm{C}-T_{water freeze}\,^{\circ}\mathrm{C}) \\
=32\,^{\circ}\mathrm{F}+\frac{212\,^{\circ}\mathrm{F}-32\,^{\circ}\mathrm{F}}{100\,^{\circ}\mathrm{C}-0\,^{\circ}\mathrm{C}}(37\,^{\circ}\mathrm{C}-0\,^{\circ}\mathrm{C})=98.6\,^{\circ}\mathrm{F}$$
(that is just to illustrate conversion it is a common but questionable human temperature)

common reference temperatures include
absolute zero
freezing point water
triple point water
human body temperature
boiling point water

when we work in absolute temperature we have absolute zero=0
so
$$T^\prime =T_A ^\prime + \frac{T_B ^\prime-T_A ^\prime}{T_B -T_A }(T-T_A)$$
becomes
$$T^\prime =0 + \frac{T_B ^\prime-0}{T_B -0 }(T-0)=\frac{T_B ^\prime}{T_B }T$$

 

[whoareyou]

K and °R are absolute temperature scales so conversion can be done by ratios. Absolute temperature can always be regarded as a change, that is absolute temperature=Absolute temperature-temperature absolute zero. °F and °C are not absolute so conversion requires also addition and subtraction. Temperature conversion in general can be effected by two reference temperatures say A and B
$$T^\prime =T_A ^\prime + \frac{T_B ^\prime-T_A ^\prime}{T_B -T_A }(T-T_A)$$
primes denote measurement in the new system
example
$$T_{human} (\,^{\circ}\mathrm{F}) =T_{water freeze} (\,^{\circ}\mathrm{F}) + \frac{T_{water boil} \,^{\circ}\mathrm{F}-T_{water freeze}\,^{\circ}\mathrm{F} }{T_{water boil}\,^{\circ}\mathrm{C} -T_{water freeze} \,^{\circ}\mathrm{C}}(T_{human}\,^{\circ}\mathrm{C}-T_{water freeze}\,^{\circ}\mathrm{C}) \\
=32\,^{\circ}\mathrm{F}+\frac{212\,^{\circ}\mathrm{F}-32\,^{\circ}\mathrm{F}}{100\,^{\circ}\mathrm{C}-0\,^{\circ}\mathrm{C}}(37\,^{\circ}\mathrm{C}-0\,^{\circ}\mathrm{C})=98.6\,^{\circ}\mathrm{F}$$
(that is just to illustrate conversion it is a common but questionable human temperature)

common reference temperatures include
absolute zero
freezing point water
triple point water
human body temperature
boiling point water

when we work in absolute temperature we have absolute zero=0
so
$$T^\prime =T_A ^\prime + \frac{T_B ^\prime-T_A ^\prime}{T_B -T_A }(T-T_A)$$
becomes
$$T^\prime =0 + \frac{T_B ^\prime-0}{T_B -0 }(T-0)=\frac{T_B ^\prime}{T_B }T$$

I'm kinda slow ... lol. How does this relate to my example (the example above)?

 

[lurflurf]

Your example pertains to temperature conversion. That is how temperature conversion works. In particular that absolute temperatures can be converted by using ratios. If the temperature is not absolute only differences can be converted by ratios. Units can be converted by ratios when 0 is the same point for all units.

 

[whoareyou]

So let me see if I understand, conversion between absolute temperatures can be done with a ratio because both scales start at the same point (ie. 0) and conversion between non-absolute scales can only be done with temperature differences since those scales aren't aligned.

So then in that same example slide that I posted previously, if you were to convert °R to °C, the final answer would have to be Δ°C and not °C?

 

[lurflurf]

That is pretty much it. To define a temperature scale we fix a zero and a size of unit. This is different than most units where we fix the same zero. Note that °C and K have units the same size and °R and °F do as well. I do not think Δ°C is standard notation, but it makes clear we are talking about the size of the unit and not its actual value. It might be best to only use absolute temperature units in R, the bigger problem is not R but the fact the equations R is used in would need to be changed. For example the ideal gas law pv=nRT would need to be written pv=nR(T-T(absolute zero))
the temperature unit in R would them be the size of the temperature used. In this case we in fact have a temperature difference. Absolute temperatures can always be regarded as differences as T=T-T(absolute vero)=T-0