a hollow alu8minum cylinder 20.0 cm deep has an internal capacity of 2.000 l at

Question

a hollow alu8minum cylinder 20.0 cm deep has an internal capacity of 2.000 l at 20.0 degrees c. It is completely filled with turpentine at 20.0 degrees c. The turpentine and the aluminum cylinder are then slowly warmed together to 80.0 degrees c. How much turpentine overflows? What is the volume of the turpentine remaining in the cylinder at 80.0 degrees c? If the combination with this amount of turpentine is then cooled back to 20.0 degrees c, how far below the cylinder's rim does the turpentine's surfacde recede?

Explanation / Answer

We assume, before heating, that the aluminum cylinder and the turpentine are both at temperature 20.0 C. (Otherwise, we don't have enough information to solve the problem.) Then both are heated to 80.0 C, a temperature increase of 60 C. The linear dimensions of the aluminum each expand by the factor (1 + 60*(24e-6) = 1 + 1.44e-3 , so the volume expands by that factor cubed: V' = V*(1 + 1.44e-3)^3 = 1.004326 * V But V = 2 L so V' = 2.00865 L. In the meantime, the turpentine has expanded by factor 1 + 60*9e-4 = 1 + 5.4e-2, so from a volume of 2 L to 2*(1.054) = 2.108 L. Therefore, the amount of overflow is 2.108 - 2.00865 = 0.09935 L 2) the volume remaining in the cylinder is that of the cylinder .. 2.00865 L 3) When everything has cooled down to 20.0 C again, the issue is: How much turpentine did we lose? Well, the overflow was 0.09935 L out of what should have been a total of 2.108 L. This is: .04713 or about 4.7%. So when all is back to normal temperature, the cylinder should be filled by 4.7% less: Therefore, the reduction in filling level should be: 0.04713 * 20 cm = 0.9426 cm.

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