How is temperature measured using thermometer? 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
 A: 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.
A: 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.
