Why do temperature rate conversions drop the offset constant?

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transmini
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Why is that whenever we convert a temperature, say ##^\circ C## to ##^\circ F##, we use ##T_F = \frac{9}{5}T_C+32##, but whenever it involves a rate, say ##^\circ C/m##, we only multiply by the slope of the previously given formula?

I would say it's because of the latter being a rate so whenever we increase by ##5^\circ C## we increase by ##9^\circ F##, but similar conversions like distance and distance\time use the same conversion factor despite being a number vs a rate.
 

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transmini said:
Why is that whenever we convert a temperature, say ##^\circ C## to ##^\circ F##, we use ##T_F = \frac{9}{5}T_C+32##, but whenever it involves a rate, say ##^\circ C/m##, we only multiply by the slope of the previously given formula?
A rate doesn't have an origin.
I would say it's because of the latter being a rate so whenever we increase by ##5^\circ C## we increase by ##9^\circ F##, but similar conversions like distance and distance\time use the same conversion factor despite being a number vs a rate.
I don't think that's true.
 
russ_watters said:
A rate doesn't have an origin.

I don't think that's true.

What part don't you think is true?
 
transmini said:
What part don't you think is true?
Neither distance nor speed use an origin when converting...

...though for distance it is because you use the same origin regardless of the units.
 
transmini said:
Why is that whenever we convert a temperature, say ##^\circ C## to ##^\circ F##, we use ##T_F = \frac{9}{5}T_C+32##, but whenever it involves a rate, say ##^\circ C/m##, we only multiply by the slope of the previously given formula?
What do you mean? ##100^\circ C/m = 212^\circ F/m \neq \frac{9}{5} \cdot 100^\circ C/m = 180^\circ C/m##
 
fresh_42 said:
What do you mean? ##100^\circ C/m = 212^\circ F/m \neq \frac{9}{5} \cdot 100^\circ C/m = 180^\circ C/m##

Well that's my point, they aren't equal. In every problem I've been given we use the latter method. As an example, the second paragraph under the heading Moist Adiabatic Lapse Rate on this article: Lapse Rate
It lists the dry adiabatic lapse rate as being ##9.8^\circ C/km## or equivalently ##5.38^\circ F/1000ft##
to get to there we would do ##\frac{9.8^\circ C}{1 km}*\frac{9^\circ F}{5^\circ C}*\frac{.3048 km}{1000 ft} = \frac{5.38^\circ F}{1000 ft}##

But initially I would've assumed to use the formula for Celsius to Fahrenheit instead of just the rate
 
The point is not the formula. The point is that the adiabatic rate is a difference of temperatures, not an absolute value. Thus
$$
(100-50)°C=((\frac{9}{5}100 +32)°F-(\frac{9}{5} 50 +32)°F)=((\frac{9}{5} 100)°F-(\frac{9}{5} 50)°F)= \frac{9}{5} (100-50)°F
$$
The difference makes the translation term of the affine transformation cancel out, not the rate.
 
Strictly, a temperature in °C is a point on a scale, not a quantity, so you should give the lapse rate as 9.8 K/km, not °C. 9.8°C is a specific temperature, equal to 282.95 K. You would presumably not make the mistake of quoting the lapse rate as 283 K/km! But people often do that with C and F, e.g. in a newspaper article you may read of a temperature of "-10°C (-50°F)" because someone has used a converter to find that 10°C = 50°F. Or that the temperature at the top of a 1000m mountain is "10°C (50°F)" cooler than at the bottom. C and F temperatures are both points on a scale, but the zeros of the two scales are different, hence the constant of 32 in the conversion equation. When you are concerned with temperature differences, however, it is only the slope of the equation that matters.