Columns of liquids at different temperuatures

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A problem involving two vertical columns of liquid at different temperatures is presented, where a 50 cm column at 50°C balances a 60 cm column at 100°C. The calculations show that the density relationship leads to a coefficient of cubical expansion of 0.004 °C, while the book states it should be 0.005 °C. The discussion highlights the importance of considering significant changes in volume and the complexities of volumetric thermal expansion. A suggestion is made to use the definition of the volumetric thermal expansion coefficient for clarity. The conversation emphasizes the need for accurate calculations in thermal expansion problems.
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Sir,
Please help me with this problem.
# A vertical column of liquid 50 cm long at 50 degree Celsius balances another column of same liquid 60 cm long at 100 degree Celsius. What is the coefficient of absolute expansion of the liquid?
I solved it in the following way:
Pressure exerted by 50 cm of liquid at 50 degree Celsius = Pressure exerted by 60 cm of liquid at 100 degree Celsius.
Hence, (h1)(d1)g = (h2)(d2)g
50 x d1 x g = 100 x d2 x g
d1 = 2 x d2 -------- (1)
We know that d1 = d2[1 + r(dt)] --------- (2)
Where d1 = density at T1 temperature
d2 = density at T2 temperature(Here T1<T2)
r = coefficient of cubical expansion of the liquid
dt = T2 – T1
Substituting (1) in (2) we get,
2 x d2 = d2[1 + (r x 50)]
By solving we get,
r = 0.004 / deg Celsius
But the answer given in my book is 0.005/ deg Celsius. Please advice.
 
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Please respond.
 
Amith2006 said:
Pressure exerted by 50 cm of liquid at 50 degree Celsius = Pressure exerted by 60 cm of liquid at 100 degree Celsius.
Hence, (h1)(d1)g = (h2)(d2)g
50 x d1 x g = 100 x d2 x g
h2=60 (not 100) cm
 
Amith2006 said:
Sir,
Please help me with this problem.
# A vertical column of liquid 50 cm long at 50 degree Celsius balances another column of same liquid 60 cm long at 100 degree Celsius. What is the coefficient of absolute expansion of the liquid?
I solved it in the following way:
Pressure exerted by 50 cm of liquid at 50 degree Celsius = Pressure exerted by 60 cm of liquid at 100 degree Celsius.
Hence, (h1)(d1)g = (h2)(d2)g
50 x d1 x g = 100 x d2 x g
d1 = 2 x d2 -------- (1)
We know that d1 = d2[1 + r(dt)] --------- (2)
Where d1 = density at T1 temperature
d2 = density at T2 temperature(Here T1<T2)
r = coefficient of cubical expansion of the liquid
dt = T2 – T1
Substituting (1) in (2) we get,
2 x d2 = d2[1 + (r x 50)]
By solving we get,
r = 0.004 / deg Celsius
But the answer given in my book is 0.005/ deg Celsius. Please advice.
Though I typed it wrongly, I solved it by using h2 = 60cm. Is it right?
 
Amith2006 said:
Though I typed it wrongly, I solved it by using h2 = 60cm. Is it right?
The problem here is that the change in volume is significant. So the rate of change in volume increases as the volume and temperature increase, does it not? It is a little more complicated than simple linear expansion.

AM
 
Last edited:
Amith, you need to use :

\gamma = \frac{1}{V} \frac{\partial V}{\partial T}

This is the definition of volumetric thermal expansion coefficient.
 
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