I know that there is mercury present inside a mercury thermometer and it expands or contracts depending on temperature of object we are measuring. Now I don't understand why thermal equilibrium plays a role in this. (I read it in a book and I have no idea why is this true?). I think as follows: Say a body has temperature of 40°c, now say temperature of mercury in the thermometer is around 10°c, so thermal transfer happens until there is thermal equilibrium. But I don't see it giving the required temperature of object. I request the answer to my problem


2 Answers 2


The scale on a mercury thermometer is actually measuring the amount by which the mercury in the thermometer has expanded, which can in turn be used to find the temperature of the mercury (the scale, which will be labelled in degrees centigrade and/or degrees Fahrenheit, does this conversion for you).

If the mercury in the thermometer is not in thermal equilibrium then it is exchanging heat with the outside environment, so the reading from the thermometer will be changing. However, if you wait until the thermometer reaches thermal equilibrium then you know it is at the same temperature as whatever it is in contact with, so by measuring the temperature of the mercury you are also measuring the temperature of the other object.

  • $\begingroup$ So, isn't the temperature measured by thermometer same as body's temperature? $\endgroup$
    – RAHUL
    May 19, 2021 at 14:02
  • 1
    $\begingroup$ @RAHUL The thermometer is measuring its own temperature. This is only the same as the temperature of whatever it is in contact with once it has reached thermal equilibrium. $\endgroup$
    – gandalf61
    May 19, 2021 at 14:58

Theoretically, you are correct, when the thermometer does reach thermal equilibrium, the temperature it measures would be different from the initial temperature of the object. However, mercury has a very low specific heat constant. This means that the temperature of mercury rises quickly while absorbing a small amount of heat. Moreover, the mass of mercury in the thermometer is also usually much less than the mass of the object whose temperature we measure. If $m,s,T_{Hg}$ are these values for mercury, and $M, c,T_i$ are these values for our object, then, for thermal equilibrium at $T_0$, we have: $$Mc(T_i-T_0)=ms(T_0-T_{Hg})$$ Let $Mc=P$ and $ms=S$. Then we have, $$T_0=\frac {PT_i+ST_{Hg}}{P+S}$$ Now, notice that $P>>S$, so $$T_0=\frac {T_i+{\frac SP}T_{Hg}}{1+\frac SP}\approx T_i$$

So, the change in temperature is very less, usually.

Caution: These calculations have been made to illustrate my point. In the real world, the object and thermometer would not form an isolated system and hence energy conservation cannot be applied, as there will be loss in energy. However, I still feel it serves as a good aid in my explanation.

  • 2
    $\begingroup$ "mercury has a very low specific heat constant"... the heat capacity of mercury is only low with respect to mass, but given that it has a very high density of 13.5 g/cm^3, this amounts to a pretty high heat capacity per volume. So what actually matters is the small volume of mercury in the thermometer. $\endgroup$
    – oliver
    May 19, 2021 at 13:29
  • $\begingroup$ That does not make much sense, because heat transfer is related to the masses of the bodies, and not the volume. $\endgroup$ May 19, 2021 at 13:40
  • $\begingroup$ You should rather say that it does not make much sense to you. $\endgroup$
    – oliver
    May 19, 2021 at 13:56
  • $\begingroup$ Sure, that. I should also say that it does not make $any$ sense. Ever held a thermometer? How much does it weigh? What fraction of that weight is due to the glass? $\endgroup$ May 19, 2021 at 14:51
  • $\begingroup$ What part of my implications do you have difficulties making sense of? That mercury has a pretty high density? That total heat capacity is determined by the product of (low) specific heat capacity times (high) density times volume? That the product of specific heat capacity and density is not particularly small for mercury compared to other materials? That this only leaves volume as the determinant of low total heat capacity of a piece of mercury? $\endgroup$
    – oliver
    May 19, 2021 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.