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I have been studying and there is a question in my book asking:

A faulty thermometer reads freezing and boiling point of a liquid as -5°C and 95°C respectively on the Celsius scale. What is the correct value of temperature when it reads 60°C on the faulty thermometer?

(a) 60°C (b) 65°C (c) 64°C (d) 62°C

The solution on back of the textbook is as follows:

(b) 65°C

For all temperature scales,

(X - LFP)/(UFP - LFP) = constant

where LFP is Lower fixed point, and UFP is Upper fixed point

[X - (-5)]/[95 - (-5)] = (C - 0)/(100 - 0)

=> (60 + 5)/(95 + 5) = C/100

=> C = 65°C

In this, I do not understand from where the C - 0 and 100 - 0 came in step 2

EDIT: Fixed typo from -95°C to 95°C

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    $\begingroup$ It seems it is not only a faulty termometer, but also faulty assignement. there should be -5 deg C and +95 deg C, with the correction of adding +5 deg C, probably due accidental shifting of the scale bar. Therefore 60->65, as the liquid is obviously water. C-0 and 100-0 comes from mapping function -5..+95 to 0..100. $\endgroup$
    – Poutnik
    Commented Aug 22, 2023 at 12:54
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    $\begingroup$ I think you're getting "the lamest title of the day" prize... but the day's still young here ;) $\endgroup$
    – Mithoron
    Commented Aug 22, 2023 at 13:52
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    $\begingroup$ Please change the title of question to reflect what is being asked in the body of the post. $\endgroup$
    – Nilay Ghosh
    Commented Aug 22, 2023 at 14:15
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    $\begingroup$ I’m voting to close this question because it contains typos that should be fixed (or if the textbook problem had the typo, include the source). In fact, always include the source if you cite something. $\endgroup$
    – Karsten ♦
    Commented Aug 22, 2023 at 16:10
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    $\begingroup$ @Karsten you mean the textbook name and version? $\endgroup$
    – Dadá
    Commented Aug 27, 2023 at 8:53

1 Answer 1

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Indeed a faulty question and data as pointed out by Poutnik. Instead of "a" liquid, it should be water. Don't use plug and chug formulae. When it is known that the liquid is water, recognize the fact that the freezing point is -5 °C, and the +95 °C for the boiling point on this thermometer. This statement is to show that the systematic error is constant in this temperature range. What do you have to do to make -5 °C to 0 °C and +95 °C to 100 °C? Add +5 °C. Make this correction with this understanding in observed 60 °C and you get the textbook answer.

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  • $\begingroup$ The question is nonsense, Chemically to assume a liquid is water is foolish, dangerous, and potentially fatal. A typo is the responsibility of the typist not the reader. The scale is inverted so a 60 degree reading is colder than the FP T; no listed answer can be correct. All that is needed to fix the question is one defined Kelvin temperature measurement [the BP of water was -95 at ~600Torr. =368 K or the ice point 273K less variable] That with absolute zero defines the absolute scale, a translation defines the new scale. The student who figures it out learns something. $\endgroup$
    – jimchmst
    Commented Aug 22, 2023 at 17:07
  • $\begingroup$ @jimchmst Other cases than constant offset are highly improbable, so take it as educated guess. $\endgroup$
    – Poutnik
    Commented Aug 24, 2023 at 12:06
  • $\begingroup$ But the question does not mention that the liquid is water $\endgroup$
    – Dadá
    Commented Aug 26, 2023 at 4:01
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    $\begingroup$ @InfoDaneMent, That is why we said that it was a poorly designed question by the question author (not you!). $\endgroup$
    – ACR
    Commented Aug 26, 2023 at 4:17

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