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

A solid metal block of mass 4.0 kg4.0\text{ kg} has a density of 8000 kg m−38000\text{ kg m}^{-3} at 20∘C20^\circ\text{C}. If the linear expansivity of the metal is 2.0×10−5 K−12.0 \times 10^{-5}\text{ K}^{-1}, calculate the increase in volume of the block, in cm3\text{cm}^3, when its temperature is raised to 120∘C120^\circ\text{C}.

Answer: 3 cm^3

Answer

The increase in volume of the block is 3.0 cm33.0\text{ cm}^3.
First, the initial volume of the metal block is found using V0=mρ=4.0 kg8000 kg m−3=5.0×10−4 m3=500 cm3V_0 = \frac{m}{\rho} = \frac{4.0\text{ kg}}{8000\text{ kg m}^{-3}} = 5.0 \times 10^{-4}\text{ m}^3 = 500\text{ cm}^3. Next, because a solid expands in three dimensions, the cubic expansivity is γ=3α=3×(2.0×10−5 K−1)=6.0×10−5 K−1\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}. The temperature change is ΔT=120∘C−20∘C=100 K\Delta T = 120^\circ\text{C} - 20^\circ\text{C} = 100\text{ K}. Finally, the increase in volume is computed as ΔV=V0γΔT=500 cm3×(6.0×10−5 K−1)×100 K=3.0 cm3\Delta V = V_0 \gamma \Delta T = 500\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 3.0\text{ cm}^3.

Step-by-Step Solution

1
Calculate the initial volume of the metal block from its mass and density.
V0=mρ0=4.0 kg8000 kg m−3=5.0×10−4 m3=500 cm3V_0 = \frac{m}{\rho_0} = \frac{4.0\text{ kg}}{8000\text{ kg m}^{-3}} = 5.0 \times 10^{-4}\text{ m}^3 = 500\text{ cm}^3
Volume is equal to mass divided by density.
2
Convert linear expansivity (α\alpha) to volume expansivity (γ\gamma).
γ=3α=3×(2.0×10−5 K−1)=6.0×10−5 K−1\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}
For an isotropic solid, cubic expansivity is three times its linear expansivity.
3
Determine the change in temperature.
ΔT=120∘C−20∘C=100 K\Delta T = 120^\circ\text{C} - 20^\circ\text{C} = 100\text{ K}
Thermal expansion is driven by the change in temperature.
4
Calculate the total volume expansion.
ΔV=V0γΔT=500 cm3×(6.0×10−5 K−1)×100 K=3.0 cm3\Delta V = V_0 \gamma \Delta T = 500\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 3.0\text{ cm}^3
Applying the thermal volume expansion formula ΔV=V0γΔT\Delta V = V_0 \gamma \Delta T.

Key Concept

Thermal Volume Expansion of Solids
Estimated Time:2m 0s
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Topics

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

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