Thermal Expansion: Solve Problem w/Turpentine & Al Cylinder

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SUMMARY

The discussion centers on solving a thermal expansion problem involving a hollow aluminum cylinder and turpentine. The cylinder, with a depth of 18.0 cm and an internal capacity of 2.000 L at 24.0°C, is filled with turpentine and heated to 88.0°C. The user correctly identifies that both the cylinder and the turpentine expand, but the turpentine overflows due to its greater expansion. The user seeks confirmation on their approach to calculating the remaining volume of turpentine after heating.

PREREQUISITES
  • Understanding of thermal expansion principles
  • Familiarity with volume calculations for cylindrical shapes
  • Knowledge of the thermal expansion coefficients for liquids
  • Basic algebra for solving equations
NEXT STEPS
  • Research the thermal expansion coefficient of turpentine
  • Learn how to calculate the volume of a cylinder at varying temperatures
  • Study the concept of overflow in thermal expansion scenarios
  • Explore the use of Table 19.1 for thermal expansion calculations
USEFUL FOR

Students studying thermodynamics, engineers working with thermal systems, and anyone involved in fluid mechanics or material science.

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Homework Statement



A hollow aluminum cylinder 18.0 cm deep has an internal capacity of 2.000 L at 24.0°C. It is completely filled with turpentine and then slowly warmed to 88.0°C. Note: If applicable, Table 19.1 is available for use in solving this problem.

The Attempt at a Solution



I solved part A and got an answer.

I can't figure out what I am doing wrong for part B, maybe someone can help.

When the cylinder heated up it expanded, as did the turpentine; the turpentine expanded more and then overflowed. So the remaining turpentine is the volume of the cylinder at 88C. So here is what I have set up.

Ok, it will take me a while to get the math in here, so maybe someone can just tell me if my concept is wrong and then that will be easy.

I know the volume of turpentine is the volume of the cylinder at 80C. So, with that, I can refind the volume of turpentine at 24C. Once I have the volume of the turpentine at 24C then I can set that to pi r^2 h, and solve for h. I can get r from initial conditions I was given. What is wrong with this thought process?
 
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If I need to put the math tell me and I will do it.
 
Ohhhhhhhh nothing is wrong with that thought process is it? Thats right, someone tell me that right.
 

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