Question icon
Grade 11Thermal Physics

A hole is drilled in a copper sheet. The diameter of the hole is 4.24cm. at 270c.What iis the change in the diameter of the hole when sheet is heated to 2270c?
(coeff. of linear expansion of copper is 1.70x10-5 0c-1

Profile image of Hermionie
9 Years agoGrade 11
Answers icon

1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer0 Years ago

To determine the change in the diameter of the hole in the copper sheet when it is heated from 27°C to 227°C, we can use the concept of linear expansion. The formula for linear expansion is quite straightforward and can be applied here.

Understanding Linear Expansion

Linear expansion refers to the increase in length (or diameter, in this case) of a material when it is heated. The formula for calculating the change in length due to temperature change is:

ΔL = L₀ × α × ΔT

  • ΔL = change in length (or diameter)
  • L₀ = original length (or diameter)
  • α = coefficient of linear expansion
  • ΔT = change in temperature

Applying the Formula

In this scenario:

  • The original diameter of the hole, L₀, is 4.24 cm.
  • The coefficient of linear expansion for copper, α, is 1.70 x 10-5 °C-1.
  • The change in temperature, ΔT, is 227°C - 27°C = 200°C.

Calculating the Change in Diameter

Now, we can plug these values into the formula:

ΔL = 4.24 cm × (1.70 x 10-5 °C-1) × 200°C

Calculating this step-by-step:

  • First, calculate the product of the coefficient of linear expansion and the temperature change:
  • 1.70 x 10-5 × 200 = 3.4 x 10-3

  • Next, multiply this result by the original diameter:
  • ΔL = 4.24 cm × 3.4 x 10-3 = 0.014416 cm

Final Diameter Calculation

The change in diameter of the hole is approximately 0.0144 cm. To find the new diameter of the hole after heating, we add this change to the original diameter:

New Diameter = Original Diameter + ΔL

New Diameter = 4.24 cm + 0.0144 cm = 4.2544 cm

Summary

When the copper sheet is heated from 27°C to 227°C, the diameter of the hole increases from 4.24 cm to approximately 4.2544 cm due to thermal expansion. This example illustrates how materials respond to temperature changes and the importance of understanding thermal properties in practical applications.