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3 years ago

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A spherical steel ball bearing has a diameter of 2.540 cm at 26.00°C. (Assume the coefficient of linear expansion for steel is 1

1 ✕ 10−6 (°C)−1. )
(a) What is its diameter when its temperature is raised to 91.0°C? (Give your answer to at least four significant figures.) 165.1 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm
(b) What temperature change is required to increase its volume by 1.100%

1 answer:

Lunna [17]3 years ago

7 0

Answer: a)2.542cm

Explanation:

According to area expansivity which is defined as change in area per unit area for degree rise in kelvin.

Area expansivity= A2-A1/A1(¶2-¶1)

A2-A1 is change in area

¶2-¶1 is temperature change

A2 if final area

A1 is initial area

¶2 is final temp = 91°C

¶1 is initial temp= 26°C

coefficient of linear expansion for steel is 11 ✕ 10−6 (°C)−1.

Area of the spherical steel ball = Πd²/4

A1= Π×2.54²/4

A1 = 5.07cm²

Area expansivity = 2×linear expansion = 2×11 ✕ 10−6 (°C)−1.

= 22 ✕ 10−6 (°C)−1.

Substituting in the formula to get final area A2

22 ✕ 10−6 (°C)−1 = A2-5.07/5.07(91-26)

22 ✕ 10−6 (°C)−1 = A2-5.07/329.55

A2-5.07 = 0.0073

A2 = 0.0073+5.07

A2= 5.0073cm²

To get final diameter

A2=Πd²/4

5.0073=Πd²/4

20.309 = Πd²

d² = 20.309/Π

d²=6.46

d= √6.46

d= 2.542cm

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