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A square hole 5.3 cm along each side is cut in a sheet of copper. calculate the change in the area of this hole if the temperature of the sheet is increased by 23 k. the coefficient of linear expansion of copper is 1.7 × 10−5 ( ◦c)−1 .

User Menaka
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Final answer:

The change in the area of a square hole in a sheet of copper caused by a temperature increase of 23K is approximately 0.032 cm², by applying the coefficient of linear expansion for copper.

Step-by-step explanation:

To calculate the change in the area of the square hole in the copper sheet with a temperature increase, we need to apply the principles of thermal expansion. The coefficient of linear expansion (α) for copper is given as 1.7 × 10⁻µ (°C)⁻¹. Since area expansion is involved, the change in each dimension due to temperature is effectively doubled, leading to the area changing by a factor of the linear expansion coefficient times two.



The original area (A1) of the square hole is:


  • A1 = side × side = 5.3 cm × 5.3 cm = 28.09 cm²


The change in length (ΔL) of a side due to thermal expansion is calculated by ΔL = αLΔT, where:


  • α = 1.7 × 10⁻µ (°C)⁻¹ (coefficient of linear expansion)

  • L = original length of the side (5.3 cm)

  • ΔT = change in temperature (23 K or °C)

Substituting the values, we get:

ΔL = 1.7 × 10⁻µ × 5.3 cm × 23 = 0.0020889 cm per side.

The new length of each side is:

L + ΔL = 5.3 cm + 0.0020889 cm = 5.3020889 cm

The new area (A2) of the square hole is:

A2 = (5.3020889 cm)² = approximately 28.122 cm²

The change in the area (ΔA) of the hole is:

ΔA = A2 - A1 = approximately 28.122 cm² - 28.09 cm² = approximately 0.032 cm²

So, the change in the area of the square hole when the temperature of the copper sheet is increased by 23 K is approximately 0.032 cm².

User Lesmian
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