Question

A hollow aluminum cylinder 21.0 cm deep has an internal capacity of 2.000 L at 25.0°C. It is completely filled with turpentine at 25.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 77.0°C. (The average linear expansion coefficient for aluminum is 24 ✕ 10−6°C−1, and the average volume expansion coefficient for turpentine is 9.0 ✕ 10−4°C−1.)

(a) How much turpentine overflows? cm3

(b) What is the volume of turpentine remaining in the cylinder at 77.0°C? (Give your answer to at least four significant figures.) L

(c) If the combination with this amount of turpentine is then cooled back to 25.0°C, how far below the cylinder's rim does the turpentine's surface recede? cm

Answer 1

(a) Change in temperature, T = (77-25) = 52^oC

Final volume of cylinder, V_f = V_i(1+3T) = 2*(1+3*24*10^-6*52) = 2.0074 L

Final volume of turpentine, V_f = V_i(1+T) = 2*(1+9*10^-4*52) = 2.0936 L

So, volume of turpentine overflown = (2.0936-2.0074) L. = 0.0862 L = (0.0862*1000) = 86.2 cm³

(b) Volume of turpentine remaining = Final volume of cylinder. = 2.0074 L

(c) When cooled back to 25^oC, the volume of turpentine decreases from 2.0074 L to a final value, V_f = V_i(1+T)

So, V_f = 2.0074[1+9*10^-4*(-52)] = 1.913 L

Now, Volume of turpentine/volume of aluminium = height turpentine/height of aluminum

Since, volume = Area of cross section*length

Thus, 1.913/2 = h_turpentine/h_aluminium = h_{turpentine/21}

So, h turpentin = 20.087 cm

Hence, required distance of reced = (21-20.087) = 0.913 cm .