A hollow aluminum cylinder 25.0 cm deep has an internal capacity of 2.000 L at 1

Question

A hollow aluminum cylinder 25.0 cm deep has an internal capacity of 2.000 L at 15.0°C. It is completely filled with turpentine at 15.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 67.0°C. (The average linear expansion coefficient for aluminum is 2.4 10-5°C1, and the average volume expansion coefficient for turpentine is 9.0 10-4°C1.)

(a) How much turpentine overflows?
cm3

(b) What is the volume of turpentine remaining in the cylinder at 67.0°C? (Give your answer to four significant figures.)
cm3

(c) If the combination with this amount of turpentine is then cooled back to 15.0°C, how far below the cylinder's rim does the turpentine's surface recede?
cm

Explanation / Answer

a)We know that, increasing the temperature leads to the expansion of most solids
or liquids.
Now,
let alfa be the linear coefficient of the given substance
now,
V(final of the aluminum) = V(initial)*[3*alfa*(tf-ti)] - Vinitial
= (2*3*2.4*10^-5*(67-15))+2= 2.0075L
Also,V(final turpentine) = V(initial)*[3*alfa*(tf-ti)] + Vinitial
= 2*3*9*10^-4*(67-15) + 2= 2.2808L
So, volume of turpentine overflown= V(final turpentine)-V(final of the aluminum)
= 0.2733L= 273.3 cm^3

b) the volume of turpentine remaining in the cylinder= Final volume of the cylinder
= 2.0075 L = 2007.5 cm^3.

c) If we change the temperature to reach the original one,then the overflown amount
over the initial volume will give us
0.2733/2.0075 = 0.1361
this ration also applies to the edges as well as the volume
So,we will multiply the edge (25 cm) by the ratio (0.1361)
we will get 3.4025 cm below the surface

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