Solving Thermo Dynamics Problem: Aluminum Cylinder Filled with Turpentine

[elabed haidar]

Homework Statement

A hollow aluminum cylinder 20.0 cm deep has an internal
capacity of 2.000 L at 20.0°C. It is completely filled with
turpentine and then slowly warmed to 80.0°C. (a) How
much turpentine overflows? (b) If the cylinder is then
cooled back to 20.0°C, how far below the cylinder’s rim
does the turpentine’s surface recede? we have the coeffiecent of turpentine and aluminium just help me solve this question without necessaryly using the numbers

Homework Equations

The Attempt at a Solution

i need help actually

 

[mark726]

solving this problem.

Hello,

Thank you for your post. To solve this problem, we can use the equation for thermal expansion:

ΔL = αLΔT

Where ΔL is the change in length, α is the coefficient of thermal expansion, L is the original length, and ΔT is the change in temperature.

For part (a), we can use this equation to find the change in length of the cylinder as it is warmed from 20.0°C to 80.0°C. Since the cylinder is completely filled with turpentine, the change in length will correspond to the amount of turpentine that overflows. We can then use the known coefficient of thermal expansion for aluminum and turpentine to calculate the amount of overflow without using specific numbers.

For part (b), we can use the same equation to find the change in length as the cylinder is cooled back to 20.0°C. This will correspond to the amount that the turpentine's surface recedes below the cylinder's rim. Again, we can use the known coefficients of thermal expansion to calculate this without using specific numbers.

I hope this helps. Let me know if you have any further questions.