Thermal Expansion?

Question

With reference to my previous question:
"A 50cm silver bar becomes shorter by 1mm when cooled.How much was it cooled? coeff. of linear expn=.000018/c.
This problem was solved taking L=50cm.
My question is how can L=50cm?Should it not be the final lengthand the original length=50-.1=49.9? Since it is a question of expansion and not contraction."

Suppose a silver bar which was originally 50cm long expanded to 50.1cm and the temperature difference is 111.11deg celsius, value of alpha would be calculate as .000018.
But suppose we are given the data as follows: A silver bar 50.1 cm long reduces to 50 cm when cooled through a temperature difference of 111.11 deg celsius we get alpha as
.0000179.
So should we not take the initial length as the one at the lower temperature?

Answer 1

Change in length = (coefficient of expansion or contraction) x (Reference length ) x (difference between temperature reference and new temperature)

dL=Cl l (T1- T2)

dL=Cl L dT
the question is how much was it cooled or dt

dt=dL/(Cl L)= .1cm/(.000018 x 50)=111 (degrees C or K)

This is how much it shrunk after being cooled 111 degrees (C or K)

50cm is length at reference temperature and reference temp can be lower or higher than the final.