Question

two mercury-glass thermometer are labelled T and D, respectively. On a T scale, at 1 atmosphere, ice melts at -10T,and pure water boils at 80T. On the D scale ice melts at 20T and pure water boils at 100T at 1 atmospheric pressure.
(a) Derive an equation relating the two temperature scales.
(b) What would the temperature on D reading if on T it reads 51T?

Answer 1

I don't understand the problem.
I get the T and D scales and I understand that the T scale is -10 to 80. On the D scale, however, did you mean ice melts at 20 <b>on the D scale</b> and boils at <b>100 on the D scale?</b>

Answer 2

yes

Answer 3

OK. Draw two vertical lines parallel to each other, like this.
T...D
|...|________80T and 100D
|...|
|...|
|...|
|...|
|...|________-10T and 20D
|...|
|...|
|...|_______-32.5T and 0 D
The left line is the T scale; the right line is the D scale. Now draw three horizontal lines from left to right. The top line label as I have done above as 80 on T scale and 100 on the D scale. The middle horizontal line label -10 on the T scale and 20 on the D scale. I did the best I could on the above drawing; it isn't possible to do much better on this board.
So, notice that the T scale has 90 divisions [80-(-10)] between freezing and boiling points of water while the D scale has 80 divisions (100-20).
The lowest horizontal line is the zero line for the D scale. That will be how many divisions on the T scale? It will be 20*(90/80) = 22.5 so that will be 22.5 divisions less than the -10 or at -32.5T.
If I've not goofed, we convert from T to D by
1. add 32.5 (to make up for the difference in the zero mark on the D scale), then multiply by 80/90.
(T+32.5)(80/90) = D.
Let's try it on the two we know.
(80T+32.5)(80/90) = 100D
(-10T+32.5)(80/90) = 20D

Or the other way,
D(90/80)-32.5 = T
100(90/80)-32.5 = 80T
20(90/80)-32.5 = -10T