A hollow aluminum cylinder 19.0 cm deep has an internal capacity of 2.000 L at 23.0°C. It is completely filled with turpentine at 23.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 91.0°C. (The average linear expansion coefficient for aluminum is 2.4 x10^-5/°C, and the average volume expansion coefficient for turpentine is 9.0 x10^-4/°C.)
(a) How much turpentine overflows?
in cm^3?
(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.)
in cm^3?
(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede?
in cm?

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Respuesta :

jbiain jbiain · Answer 1 · 2020-03-07T05:04:36+01:00

Answer:

(a) 0.1134 L

(b) 2.009 L

(c) 0.07 cm from the top

Explanation:

The volume of the aluminum container and the turpentine is 2.000 L at 23 °C.

(a) When the entire system is heated to from 23 °C to 91 °C. The volume of the aluminum container is:

Vf = Vi*(1 + 3α*ΔT)

where:

Vi is the initial volume, Vf the final volume, α is the linear expansion coefficient and ΔT is the temperature change.

Vf = 2*[1 + 3*2.4 *10^-5*(91 - 23)] = 2.009 L

The volume of the turpentine is

Vf = Vi*(1 + β*ΔT)

where:

β is the volume expansion coefficient

Vf = 2*[1 + 9.0*10^-4*(91 - 23)] = 2.1224 L

The overflow is:

2.1224 - 2.009 = 0.1134 L

(b) the volume of turpentine remaining in the cylinder at 91 °C is the same volume of the hollow aluminum cylinder, that is, 2.009 L

(c) Cooling the system back to 23 °C means the aluminum container is back to 2.000 L. The volume of turpentine is:

Vf = 2.1224*[1 + 9.0*10^-4*(23-91)] = 1.9925 L

The following proportion must be satisfied:

Volume of turpentine / Volume of container = deep turpentine / deep of container

1.9925 / 2.000 = deep turpentine/19

deep turpentine = (1.9925 / 2.000)*19 = 18.93 cm

This is 19 - 18.93 = 0.07 cm from the top