Thermal Expansion Ethanol Problem

Click For Summary
SUMMARY

The discussion centers on a thermal expansion problem involving 108 cm³ of ethanol at -10.0 degrees Celsius poured into a glass graduated cylinder at 20.0 degrees Celsius. The final temperature of the ethanol upon reaching thermal equilibrium is calculated to be -0.892 degrees Celsius. Participants express confusion regarding the calculation of overflow without knowing the initial volume of the cylinder, emphasizing the need to consider both the expansion of the ethanol and the contraction of the glass cylinder, which has a coefficient of volume expansion of 1.2 x 10^-5 K^-1.

PREREQUISITES
  • Understanding of thermal equilibrium principles
  • Knowledge of volumetric expansion equations, specifically ΔV = β V0 (ΔT)
  • Familiarity with specific heat capacity concepts, particularly for glass (840 J/(kg*K))
  • Basic understanding of the properties of ethanol and its mass (0.0873 kg)
NEXT STEPS
  • Research the effects of thermal expansion in liquids and solids
  • Study the principles of thermal equilibrium in closed systems
  • Learn about the volumetric expansion coefficients of various materials
  • Explore practical applications of thermal expansion in laboratory settings
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone involved in laboratory experiments requiring precise temperature and volume calculations.

Fizzicist
Messages
53
Reaction score
0

Homework Statement



You pour 108 cm^3 of ethanol, at a temperature of -10.0 degrees C, into a graduated cylinder initially at 20.0 degrees C, filling it to the very top. The cylinder is made of glass with a specific heat of 840 J/(kg *K) and a coefficient of volume expansion of 1.2 *10^-5 K^-1; its mass is 0.110 kg. The mass of the ethanol is 0.0873 kg.

A. What will be the final temperature of the ethanol, once thermal equilibrium is reached?

(Answer: -.892 degrees C)

B. How much ethanol will overflow the cylinder before thermal equilibrium is reached?

Homework Equations



Equation for volumetric expansion:

\DeltaV = \beta V0 (\DeltaT)

The Attempt at a Solution



I honestly think this question (part B) is flawed. How could you solve this without knowing V0, the initial volume of the cylinder? And even if you did know the volume of the cylinder, you still wouldn't be able to determine the volume capacity it is able to hold (the ratio of volume capacity to volume of the cylinder varies). Is this even possible?
 
Last edited:
Physics news on Phys.org
You can do part 'a' easily.

B, You can't know how much it will overflow without knowing the volume of the cylinder.
But I think you are meant to assume that 108cc of cold ethanol will fill the cylinder to the top and it overflows when it warms - so just work out how much the ethanol expands
 
B, You can't know how much it will overflow without knowing the volume of the cylinder.
But I think you are meant to assume that 108cc of cold ethanol will fill the cylinder to the top and it overflows when it warms - so just work out how much the ethanol expands


Yeah. That seems reasonable...

I didn't like the way this question was asked. It made it sound as if you were supposed account for the contraction of the cylinder because it gave you the coefficient of volumetric expansion for the glass.
 
Yes - especially because you would need to know the shape of the cylinder!
The volume expansion of the glass isn't the volume expansion of the cylinder.
If you had a thin-walled spherical container and the glass expanded would the volume inside get bigger or would the wall just get thicker and expand outward?
 
How did you find part A?

Fizzicist said:
Yeah. That seems reasonable...

I didn't like the way this question was asked. It made it sound as if you were supposed account for the contraction of the cylinder because it gave you the coefficient of volumetric expansion for the glass.

?

You are supposed to account for the contraction of the cylinder.
 
?

You are supposed to account for the contraction of the cylinder.


TY, but I figured this out about a year ago.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
6K
Replies
2
Views
6K
Replies
8
Views
2K
Replies
6
Views
904
  • · Replies 13 ·
Replies
13
Views
21K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K