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Physics for ECE Lesson-end Activities, Assignments of Thermodynamics

Three physics problems related to volume expansion and thermal expansion. The first problem discusses the volume expansion of solid and hollow bodies made of the same material. The second problem talks about the potential problems that could occur in automobile engines with cast-iron cylinders and aluminum pistons when the engine gets too hot. The third problem involves calculating the change in length of the steel deck of the Humber Bridge in England when the temperature increases. solutions to all three problems.

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2018/2019

Available from 02/20/2022

Sethmorphothyst
Sethmorphothyst ๐Ÿ‡ต๐Ÿ‡ญ

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Physics for ECE
Lesson-end Activities
1. Two bodies made of the same material have the same external dimensions and
appearance, but one is solid and the other is hollow. When their temperature is
increased, is the overall volume expansion the same or different? Why?
- Yes, the overall volume expansion will be same in both cases. Thatโ€™s
because the bodies, being made of the same material, have the same
volume coefficient of expansion. Also, before the measurement is taken,
both objects must have come to equilibrium at the new temperature. To
further explain, the hollow one will expand the same as if it were also solid.
This is due to the molecules in which its energy increases that leads to any
two molecules of the solid to move further apart. Hence, hollow bodies
expand just like the material that surrounds them. The external
dimensions of both objects would increase the same amount.
2. Many automobile engines have cast-iron cylinders and aluminum pistons. What
kinds of problems could occur if the engine gets too hot? (The coefficient of volume
expansion of cast iron is approximately the same as that of steel.)
- With the rise in engine temperature for the iron cast and aluminum pistons,
it is expected that there would be expansion in the volume of both elements.
From the table of Coefficients of Volume Expansion in the discussion,
Aluminum has a larger thermal expansion coefficient than steel or is double
the coefficient of steel. This means that at any temperature, the diameter
of the aluminum piston will expand more than the inside diameter of the
cylinder and can cause a high heat capacity absorbed by the piston. With
this, the problems that could occur if the engine gets too hot are that the
piston will not be able to freely move inside the iron cast and it would cause
damage to engine due to the friction between the piston and the cast iron.
3. The Humber Bridge in England has the worldโ€™s longest single span, 1410 m.
Calculate the change in length of the steel deck of the span when the temperature
increases from โˆ’8โ„ƒ ๐‘ก๐‘œ 20โ„ƒ.
Solution:
โˆ†๐‘‡ = ๐‘‡ โˆ’ ๐‘‡๐‘œ
โˆ†๐‘‡ = 20โ„ƒ โˆ’ (- 8โ„ƒ)
โˆ†๐‘‡ = 28๐ถยฐ
โˆ†๐‘‡ = 28k
โˆ†๐ฟ = ๐›ผ๐ฟ0โˆ†๐‘‡
โˆ†๐ฟ = (1.2ร—10โˆ’5 ๐‘˜โˆ’1 ) (1410m) (28k)
โˆ†๐ฟ = 0.4737m
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Physics for ECE

Lesson-end Activities

  1. Two bodies made of the same material have the same external dimensions and appearance, but one is solid and the other is hollow. When their temperature is increased, is the overall volume expansion the same or different? Why? - Yes, the overall volume expansion will be same in both cases. Thatโ€™s because the bodies, being made of the same material, have the same volume coefficient of expansion. Also, before the measurement is taken, both objects must have come to equilibrium at the new temperature. To further explain, the hollow one will expand the same as if it were also solid. This is due to the molecules in which its energy increases that leads to any two molecules of the solid to move further apart. Hence, hollow bodies expand just like the material that surrounds them. The external dimensions of both objects would increase the same amount.
  2. Many automobile engines have cast-iron cylinders and aluminum pistons. What kinds of problems could occur if the engine gets too hot? (The coefficient of volume expansion of cast iron is approximately the same as that of steel.) - With the rise in engine temperature for the iron cast and aluminum pistons, it is expected that there would be expansion in the volume of both elements. From the table of Coefficients of Volume Expansion in the discussion, Aluminum has a larger thermal expansion coefficient than steel or is double the coefficient of steel. This means that at any temperature, the diameter of the aluminum piston will expand more than the inside diameter of the cylinder and can cause a high heat capacity absorbed by the piston. With this, the problems that could occur if the engine gets too hot are that the piston will not be able to freely move inside the iron cast and it would cause damage to engine due to the friction between the piston and the cast iron.
  3. The Humber Bridge in England has the worldโ€™s longest single span, 1410 m. Calculate the change in length of the steel deck of the span when the temperature increases from โˆ’8โ„ƒ ๐‘ก๐‘œ 20โ„ƒ. Solution: (^) โˆ†๐‘‡ = ๐‘‡ โˆ’ ๐‘‡ ๐‘œ โˆ†๐‘‡ = 20 โ„ƒ โˆ’ (- 8 โ„ƒ) โˆ†๐‘‡ = 28 ๐ถยฐ โˆ†๐‘‡ = 28 k

โˆ†๐ฟ = (1.2ร— 10 โˆ’^5 ๐‘˜โˆ’^1 ) (1410m) (28k) โˆ†๐ฟ = 0.4737m

  1. A U.S. penny has a diameter of 1.9 cm at 20โ„ƒ.The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is 2.6 ร— 10โˆ’5 ๐พโˆ’1. What would its diameter be on a hot day in Death Valley (60.0ยฐC)? On a cold night in the mountains of Greenland โˆ’40โ„ƒ? Solution: a. Diameter on a hot day in Death Valley Solution: b. Diameter on a cold night in the mountains of Greenland
  2. A brass rod is 190 cm long and 1.80 cm in diameter. What force must be applied to each end of the rod to prevent it from contracting when it is cooled from 130โ„ƒ ๐‘ก๐‘œ 10โ„ƒ? Solution:

โˆ†๐‘‡ = 20 k โˆ†๐ฟ = ๐›ผ๐ฟ 0 โˆ†๐‘‡ โˆ†D (change in diameter) = ๐›ผD 0 โˆ†๐‘‡ โˆ†D = (2.6ร— 10 โˆ’^5 ๐‘˜โˆ’^1 ) (1.9cm) (20k) โˆ†D = 9.88ร— 10 โˆ’^4 cm = 0.0988m D = D0 + โˆ†D 1.9cm + 9.88ร— 10 โˆ’^4 cm D = 1.901cm = 0.01901m โˆ†๐‘‡ = ๐‘‡ โˆ’ ๐‘‡๐‘œ โˆ†๐‘‡ = - 4 0.0โ„ƒ โˆ’ 20 ยฐ โˆ†๐‘‡ = - 60 ๐ถยฐ โˆ†๐‘‡ = - 60 k

๐ฟ = ๐ฟ 0 (1 + ๐›ผโˆ†๐‘‡) D = D 0 (1 + ๐›ผโˆ†๐‘‡)

D = 1.9 cm (1 + 2.6ร— 10 โˆ’^5 ๐‘˜โˆ’^1 (-60k)) D = 1.897cm = 0.01897m ๐œŽ = F A

๐œŽ = โˆ’ 9. 0 ร— 1010 ๐‘ƒ๐‘Ž (2.0ร— 10 โˆ’^5 ๐‘˜โˆ’^1 ) (10ยฐC -

130 ยฐC)

๐œŽ = โˆ’ 9. 0 ร— 1010 ๐‘ƒ๐‘Ž (2.0ร— 10 โˆ’^5 ๐‘˜โˆ’^1 ) (- 120 ยฐk) ๐œŽ = 21.6ร— 107 ๐‘ƒ๐‘Ž ๐น = ๐œŽ๐ด = 21.6ร— 107 ๐‘ƒ๐‘Ž (9ร— 10 โˆ’^3 m)^2 ๐œ‹ ๐น = 21.6ร— 107 ๐‘ ๐‘š^2

( 8. 1 ร— 10 โˆ’^5 ๐‘š^2 ) ๐œ‹

F = 54965.30507 N

A = ๐œ‹ (

๐‘‘ 2

)^2

A = ๐œ‹ (

  1. 80 ๐‘๐‘š 2

)^2

A = ๐œ‹ (0.9cm)^2 A = ๐œ‹ (9ร— 10 โˆ’^3 m)^2