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I am having problems with one of the problemsenter image description here

I have done (a) which is basically $\theta = 273.16K (19/15) = 346.00K$

For (b) I am given that the length is 15.00cm when the thermometer is incontact with water at its triple point. But I have no idea what the length is when it is in contact with the freezing point. I am also confused with the "precision" thin and how it relates to $\theta = 273.16(X/X_{tp})$

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how many cm change represents 0.01 K? What's the difference between the triple point and freezing point in k? –  Martin Beckett Sep 11 '12 at 4:18
To improve Martin's answer: what is the temperature of the freezing point ? What is the height for a given temperature T ? –  Shaktyai Sep 11 '12 at 8:30
Thank you all. I got the answer –  Hawk Sep 28 '12 at 4:28

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