We had the following thermal expansion Problems:
1)A vertical steel antenna tower is 200 m high. Calculate the change in height of the tower that takes place when the temperature changes from -20 C on a winter day to 30 C on a summer day.
2) A metal rod is 200.00 cm long at 0C and 200.18 cm long at 60C. What is its coefficient of linear expansion?
3) A metal vessel has a volume of 500.00 cm^3 at 0 C. If the coefficient of linear expansion is 0.000011 /C, what is the volume at 50 C?
4) A round hole of 6.000 cm in diameter at 0 C is cut in a sheet of brass(alpha = 0.000018 /C). Find the new diameter of the hole at 50 C.
Solutions:
(1)Given: L1 = 200m
T2 = 30 C and T1 = (-20 C) then (T2 – T1) = (30C – (-20C)) = 50 C
alpha (steel) = 0.000011/C
then: change in length is
x = (0.000011 /C) [200m/ 50C]= 0.000044 m = (L2 – L1)
(2)Given: L1 = 200.00 cm and L2 = 200.18 cm
So: x = (L2 – L1) = 0.18 cm
(T2 – T1 ) = (60 C – 0C) = 60C
Then: alpha = x/ {L1 (T2 - T1)} = 0.18 cm/ [200.00 cm (60C)]
alpha = 0.000015 /C
(3)V = 500.00 cm^3 and (T2 – T1) = 50C
alpha= is 0.000011 /C
Beta = 3(alpha) = 3( 0.000011/C) = 0.000033 /C
delta V= (500.00 cm^3)(0.000033/C) (50C) = 0.825 cm^3
then the new volume = V + delta V = 500.00 cm^3 + 0.82 cm^3 = 500.82cm^3
(4)Given: alpha = 0.000018 /C
(T2 – T1)= 50 C
D = 6.000 cm so r = 3.000 cm
And Area A = πr^2 = 28.274 cm^2
But delta A = 2 (alpha) A ((T2 – T1)
= 2 (0.000018 /C) 28.274 cm^2 (50C) = 0.051 cm^2
New Area A’ = A + delta A = 28.274 cm^2 + 0.051 cm^2
A’ = 28.325 cm^2
Then the new radius r’ = (A’/π)^ ½ = (28.325 cm^2/ 3.141)^ ½
= 3.002 cm
So the new diameter D’ = 2 r’ = 2 (3.002 cm) = 6.004 cm
2 comments:
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