# Thermal Expansion: Finding common temperature (Ring/Sphere)

• paulie
In summary, the conversation is about finding the common temperature at which the inner diameter of a steel ring and the diameter of a lead sphere are both 0.05% larger. The equations for areal and linear expansion are used to calculate the temperature, but the given values do not match the expected result. The question statement lacks accuracy, making it difficult to determine the exact answer.
paulie

## Homework Statement

Find the common temperature at which the inner diameter of the ring is 0.05% larger than the diameter of the sphere.

Given:
d = 5cm
To = 70°C
∝ = 29x10-6/C°
Steel ring
d = 4.9975cm (the sphere is 0.05% larger than the inner diameter of the ring, hence this value)
To = 70°C
∝ = 12x10-6/C°

Af = Ao(1+2∝Δt)
Δt = Tf - To

## The Attempt at a Solution

Equate both final area of sphere (S) and ring (R)
AFR = AFS (1.0005)
Substitute
Af = Ao(1+2∝Δt)
AoR(1+2∝RΔt) = (1.0005)AoS(1+2∝SΔt)
Arrange equation to find Δt
Δt = (1.0005)(AoS) - (AoR) / 2(AoR)(∝R) - (1.0005)(2)(AoS)(∝L)
Enter the values
Δt = (1.0005)(π(5cm/2)2) - (π(4.9975cm/2)2) / 2(π(4.9975cm/2)2)(12x10-6/C°) - 2(1.0005)(π(5cm/2)2)(29x10-6/C°)
Δt = -44.0416 C°
Which is incorrect because the Final temperature is supposed to be 11.2766°C, using my answer and finding for the final temperature with 70°C as initial.
Tf = 25.9584°C
EDIT:
I've found out that if I multiplied 1.0005 on the diameter of lead sphere (5cm) immediately before others, I'll get an answer of Δt = -58.6820 C° where the final temperature will be 11.3180°C which is pretty close but not the exact answer :(

In the case of ring, you have to consider linear expansion, not the areal expansion. Similarly in the care of sphere you have to consider volume expansion.

Linear expansion for both.

Nidum
@rl.bhat I tried using volume and linear but I arrived with a bigger value.

@mjc123 Tried to make both linear and I arrived with Δt = - 58.7521 C° which is pretty good, finding the final temperature I got 11.2473 °C although the given correct answer is 11.2766 °C. I'll try to recompute later.

The problem is actually one out of four question, I used area of expansion on the three others and found the exact correct answer. This is the last question and I'm having trouble in getting the exact value (11.2766 °C).

paulie said:
@rl.bhat I tried using volume and linear but I arrived with a bigger value.

@mjc123 Tried to make both linear and I arrived with Δt = - 58.7521 C° which is pretty good, finding the final temperature I got 11.2473 °C although the given correct answer is 11.2766 °C. I'll try to recompute later.

The problem is actually one out of four question, I used area of expansion on the three others and found the exact correct answer. This is the last question and I'm having trouble in getting the exact value (11.2766 °C).
The question statement violates the usual rules of specifying accuracy. E.g. it should give the lead diameter as 5.0000cm.
Having extended all the data to that accuracy, you might think that the answer can be given to that accuracy, but not so. In this question, small differences are calculated between numbers of similar magnitude. E.g. 5.0025-4.9975=0.0050, so 5 digits of accuracy falls to only two.
Thus, it does not really make any sense to quote the answer more precisely than 11°C.

But for what it is worth, I get 11.256. If you care about that difference, please show exactly how you calculated it.

gneill
That was what I was thinking but it bothered me that it was the only answer that didn't exactly gave the exact value from the answer sheet. Thanks for the help guys.

## 1. What is thermal expansion?

Thermal expansion is the tendency of matter to change in size, shape, and volume in response to a change in temperature. When an object is heated, its particles gain energy and vibrate faster, causing the object to expand. On the other hand, when an object is cooled, its particles lose energy and vibrate slower, causing the object to contract.

## 2. How does thermal expansion affect everyday objects?

Thermal expansion affects everyday objects in various ways. For example, metal objects such as bridges and train tracks can expand and contract with changes in temperature, causing them to buckle or warp. This is why expansion joints are used to allow for this movement. Additionally, substances such as water and concrete expand when frozen, which can cause damage to pipes and structures.

## 3. How is thermal expansion measured?

Thermal expansion is typically measured using a coefficient of thermal expansion (CTE). This value represents the amount of expansion or contraction that occurs per unit of length when the temperature changes by 1 degree Celsius. It is usually expressed in units of 1/°C or 1/K (Kelvin).

## 4. What is the purpose of finding a common temperature in thermal expansion?

The purpose of finding a common temperature in thermal expansion is to determine the amount of expansion or contraction that occurs in a substance when it is heated or cooled. This is important in engineering and construction, where precise measurements are necessary to ensure the stability and safety of structures.

## 5. What factors can affect thermal expansion?

Several factors can affect thermal expansion, including the material of the object, its shape and size, and the temperature change. Different materials have different coefficients of thermal expansion, meaning they will expand or contract at different rates. The shape and size of an object can also affect its expansion, as well as the direction in which it expands. Additionally, the rate of temperature change can also impact thermal expansion, with faster temperature changes resulting in larger expansions or contractions.

• Introductory Physics Homework Help
Replies
16
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
8K
• Introductory Physics Homework Help
Replies
2
Views
5K