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(I) Liquid thermometer You have a quartz thermometer with interior dimensions 0.8mm

diameter and 150 mm in height sealed at the ends with a 1mL reservoir of mercury at the

bottom. At 30 degrees the reservoir is completely filled and the tube is half

filled with mercury, and the remainder is filled with air (mostly nitrogen, ie. non-interacting).

The volumetric thermal expansion coeffcient of mercury is 182*10^6/degree

Assume the liquid is incompressible (a reasonable assumption in this situation).

(a) Find the volume of mercury in the tube. Does this seem significant compared to the

1mL in the reservoir?

(b) Find the change in volume of the mercury between 30C and 80C using just the volume

in the reservoir, and the total volume in the reservoir plus tube. Comparing these two

numbers, did the volume in the tube contribute significantly for the volume change?

(c) Find the change in length and change in diameter of the quartz tube between 30C and

80C. Use these to find the change in volume of the tube (neglecting the reservoir). Is

this signficant compared to the change in volume of the mercury?

(d) What is the height of the column of mercury at 80C?

(e) What is the remaining volume occupied by the air?

(f) If the tube was sealed at 30C (ie. the air in the tube was at atmospheric pressure at 30C), and the quartz can withstand a maximum of 150 kPa pressure, will the thermometer explode at 80C?

Explanation / Answer

(a) Find the volume of mercury in the tube. Does this seem significant compared to the 1mL in the reservoir?

V = A L = (3.1416*0.8e-3/2*0.8e-3/2) * 150e-3/2 = 3.76992e-8 m3 = 3.77 x 10^-8 m^3

==> V = 3.77x 10^-8 * 1e-6 mL = 0.0377 mL

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(b) Find the change in volume of the mercury between 30C and 80C using just the volume

in the reservoir, and the total volume in the reservoir plus tube. Comparing these two

numbers, did the volume in the tube contribute significantly for the volume change?

just the volume in the reservoir:

dV' = V0 (alpha delta(T)) = 1mL * ((182e-6)*(80-30)) = 0.0091 mL

==> dV' = 0.0091 mL

the total volume in the reservoir plus tube:

dV'' = V0 (alpha delta(T)) = (1+0.0377) * (182e-6)*(80-30) = 0.009443 mL

==> dV'' = 0.00944 mL

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(c) Find the change in length and change in diameter of the quartz tube between 30C and 80C. Use these to find the change in volume of the tube (neglecting the reservoir). Is this signficant compared to the change in volume of the mercury?

length:

dL = L0 (alpha (tf - Ti)) = 150e-3 * (0.59e-6 * (80 - 30)) = 0.000004425 m

Diameter:

dD = D0 (alpha (tf - Ti)) = 0.8e-3 * (0.59e-6 * (80 - 30)) = 0.0000000236 m

Volume:

dV = pi (D/2)^2 * L - pi (D0/2)^2 L0

==> dV = 3.1416*((0.8e-3+0.0000000236)/2)^2 * (150e-3+0.000004425) - 3.1416*(0.8e-3/2)^2 * (150e-3)

==> dV = 6.67 x 10^-12 m3

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(d) What is the height of the column of mercury at 80C?

the volume in the tube at 80C:

V = V0 (1 + alpha delta(T)) = (0.0377) * (1 + (182e-6)*(80-30)) = 0.03804307 mL = 0.03804307e-6 m3

h = V/A = V/(pi (D/2)^2) = 0.03804307e-6/(3.1416*((0.8e-3+0.0000000236)/2)^2) = 0.0756796 m

==> h = 0.0757 m = 75.7 mm

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part (e):

h' = 0.150 - 0.757 = 0.07432 m

V = h' A = 0.07432 * (3.1416*((0.8e-3+0.0000000236)/2)^2) = 0.0374 x 10^-6 m^3 = 0.0374 mL

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part (f)

Pi Vi/Ti = Pf Vf/Tf

==> Pf = (Tf/Ti) * (Vi/Vf) * Pi = (273+80)/(273+30) * ((75)/(74.32)) * (1.01e5) = 1.18e5 Pa = 118 kPa < 150 kPa

Therefore the thermometer will not explode at 80C.

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