How Is the True Temperature Calculated from a Miscalibrated Thermometer?

[poloboy]

a thermometer reads -.3 degrees celsius at ice point and 100.6 degrees celsius at steam point. find the true temperature of a thermometer if thermometer read 18.2 degree celsius

The Attempt at a Solution

this is actually a simple math question that i need a little help with. i know what the answer is because it was give in a student hand book but what i couldn't find is how the book came to the conclusions from equation (A and B) to get equation (C). If someone could point me in the mathmatical direction, it would be great

(A) 0 = A + b (-0.3)
(B) 100 = A + b (100.6)
solving A and B
Tc = A +bX = 0.2973 + 0.991X <-- how did they get A and B here?

where X = 18.2
and from equation Tc the true temperature is 18.33degrees celsius

 

[BerryBoy]

So your equations are:

$$0 = A - 0.3b$$

&

$$100 = A + 100.6b$$

You can multiply your top equation by $$\frac{100.6}{0.3}$$ and then add the two equations together :P

Let me know how you get on...
Sam

 

[poloboy]

i don't get it :|

 

[poloboy]

i think i got the answer but how come we go 100.6/0.3?

 

[GTrax]

Equations smayshions - I don't see it that way at all.
Firstly intuitively. Ice point is 0 degress C and Steam point is 100 degrees C, spanning 100 degrees.
The thermometer reads -0.3 degrees and +100.6 degrees corresponding, spanning 100.9 degrees.
A span of a real degree is only (100/100.9) of the indicated degrees.
To get it up from -0.3 to 18.2 shifted it by 18.5 indicated degrees
This moved the real degrees by 18.5*(100/100.9) = =18.335 (OK - so round it down)

The "equation" we are talking about is the straight line Tc=(100/100.9)*Ti+C ... The slope is 100/100.9
One solution is at Tc=0, where Ti=-0.3 Solve for C to get C = 0.29732408
So we have Tc = (100/100.9)*Ti +0.29732408
Input Ti=18.2, It outputs 18.33