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Difficulty: MediumThermal Expansion of Solids (Linear, Area, and Volume Expansivity)

A cylindrical metal rivet has a diameter of 2.50 cm2.50\text{ cm} at a room temperature of 25∘C25^\circ\text{C}. It needs to be inserted into a hole of diameter 2.49 cm2.49\text{ cm} in a structural frame. By how many kelvins must the rivet be cooled so that its diameter shrinks to just match the diameter of the hole? (Linear expansivity of the metal is 2.0×10−5 K−12.0 \times 10^{-5}\text{ K}^{-1}).

Answer: 200 K

Answer

The rivet must be cooled by 200 K.
Thermal expansion or contraction of a linear dimension (such as diameter) is governed by Δd=d0αΔT\Delta d = d_0 \alpha \Delta T. Substituting Δd=−0.01 cm\Delta d = -0.01\text{ cm}, d0=2.50 cmd_0 = 2.50\text{ cm}, and α=2.0×10−5 K−1\alpha = 2.0 \times 10^{-5}\text{ K}^{-1} gives −0.01=2.50×(2.0×10−5)×ΔT-0.01 = 2.50 \times (2.0 \times 10^{-5}) \times \Delta T, leading to ΔT=−200 K\Delta T = -200\text{ K}. Hence, cooling by 200 K is required.

Step-by-Step Solution

1
Calculate the required change in diameter
\Delta d = 2.49\text{ cm} - 2.50\text{ cm} = -0.01\text{ cm}
The diameter of the rivet must decrease from 2.50 cm to 2.49 cm to fit into the hole.
2
Set up the linear expansion equation
\Delta d = d_0 \alpha \Delta T
Linear contraction/expansion applies directly to any linear dimension of a solid, including diameter.
3
Substitute given values into the equation
-0.01\text{ cm} = (2.50\text{ cm}) \times (2.0 \times 10^{-5}\text{ K}^{-1}) \times \Delta T
Substitute initial diameter, linear expansivity, and change in diameter.
4
Solve for the temperature change
\Delta T = \frac{-0.01}{5.0 \times 10^{-5}} = -200\text{ K}
Dividing the change in length by the product of initial length and linear expansivity yields the temperature change.

Key Concept

Thermal Contraction and Linear Expansivity of Solids
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Topics

Thermal Expansion of Solids (Linear, Area, and Volume Expansivity)

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