Heat Energy and Expansion
Heat Energy and Expansion
Area expansivity

The area expansivity \(\beta\) of a substance is defined as the increase in area per unit area of a piece of that substance per degree rise in temperature. As heat is applied to a substance, there is an increase in its area. This increase is its area expansivity.

$$\beta = \frac{\text{change in area}}{\text{original area × Temp. rise}} $$ $$ \beta = \frac{∆A}{{A_1}{∆\theta}} $$ $$ \beta = \frac{{A_2}-{A_1}}{{A_1}({\theta_2} - {\theta_1})} $$ $$ \text{where } \beta = \text{area expansivity in }{K^{-1}} $$ $$ {A_1} = \text{area before expansion} $$ $$ {A_2} = \text{area after expansion} $$ $$ {\theta_1} = \text{Temperature before expansion} $$ $$ {\theta_2} = \text{temperature after expansion} $$ $$∆A = \text{change in length} $$ $$ ∆\theta = \text{change in temperature} $$
Volume Expansivity

The volume expansivity \(\gamma\) of a substance is defined as the increase in volume per unit volume of a piece of that substance per degree rise in temperature. It is also called cubic expansivity.

$$\gamma = \frac{\text{change in volume}}{\text{original volume × Temp. rise}} $$ $$ \gamma = \frac{∆V}{{v_1}{∆\theta}} $$ $$ \gamma = \frac{{V_2}-{V_1}}{{V_1}({\theta_2} - {\theta_1})} $$ $$ \text{where } \gamma = \text{volume expansivity in }{K^{-1}} $$ $$ {V_1} = \text{volume before expansion} $$ $$ {V_2} = \text{volume after expansion} $$ $$ {\theta_1} = \text{Temperature before expansion} $$ $$ {\theta_2} = \text{temperature after expansion} $$ $$∆V = \text{change in volume} $$ $$ ∆\theta = \text{change in temperature} $$

The relationship between linear expansivity, area and volume expansivity is given by $$ \beta = 2\alpha $$ $$ \gamma = 3\alpha $$

Calculations

Example 1: A solid metal cube of side 10cm, is heated from 10°c to 60°c. if the linear expansivity of the metal is 1.2 × 10-5K-1. Calculate the increase in its volume. (WAEC)

Solution

$$ \text{original length} = 10cm $$ $$ \text{original volume} = 10 × 10 × 10 $$ $$ = 1000cm³ $$ $$ ∆\theta = 60°C - 10°C = 50°C $$ $$ \alpha = 1.2 × {10^{-5}{K^{-1}}} $$ $$ \text{cubic expansivity }{\gamma} = 3\alpha $$ $$ \gamma = 3 × 1.2 × {10^{-5}} $$ $$ \gamma = 3.6 × {10^{-5}{K^{-1}}} $$ $$ \gamma = \frac{∆V}{{V_1}{∆\theta}} $$ $$ ∆V = \gamma × {{V_1}{∆\theta}} $$ $$ ∆V = 3.6 × {10^{-5}} × 1000 × 50 $$ $$ ∆V = 1.8cm³ $$

Example 2: A brass cube of side 10cm is heated through 30°C. If the linear expansivity of brass is 2.0 × 10-5K-1, what is the increase in its volume? (NECO)

Solution

$$ ∆V = ? $$ $$ {V_1} = 10 × 10 × 10 = 1000cm³ $$ $$ ∆\theta = 30°C $$ $$ \gamma = 3 × 2.0 × {10^{-5}} $$ $$ \gamma = 6 × {10^{-5}{K^{-1}}} $$ $$ ∆V = \gamma × {V_1} × ∆\theta $$ $$ ∆V = 6 × {10^{-5}} × 30 × 1000 $$ $$ ∆V = 1.8cm³$$

Example 3: The length of a side of a metallic cube at 20°C is 5cm. Given that the linear expansivity of the metal is \(4.0 × 10^{-5}{K^{-1}}\), find the volume of the cube at 120°C. (JAMB)


Solution

$$ {V_1} = 5 × 5 × 5 = 125cm³ $$ $$ {\theta_1} = 20°C $$ $$ {\theta_2} = 120°C $$ $$ ∆\theta = 120 -20 = 100°C $$ $$ {V_2} = ? $$ $$ \alpha = 4.0 × {10^{-5}{K^{-1}}} $$ $$ \gamma = 3 × 4.0 × {10^{-5}} $$ $$ \gamma = 12 × {10^{-5}{K^{-1}}} $$ $$ {V_2} = \gamma × {V_1} × {∆\theta} + {V_1} $$ $$ {V_2} = 12 × {10^{-5}} × 125 × 100 + 125 $$ $$ {V_2} = 126.5cm³ $$

Example 4: A blacksmith heated a metal whose cubic expansivity is \(6.3 × {10^{-6}{K^{-1}}}\). The area expansivity is ?. (JAMB)

Solution

$$ \gamma = 6.3 × {10^{-6}{K^{-1}}} $$ $$ \gamma = 3\alpha $$ $$ \beta = 2\alpha $$ $$ \therefore \alpha = \frac{\gamma}{3} $$ $$ \alpha = \frac{6.3 × {10^{-6}}}{3} $$ $$ \alpha = 2.1 × {10^{-6}{K^{-1}}} $$ $$ \therefore \beta = 2.1 × {10^{-6}} × 2 $$ $$ \text{Area expansivity} = 4.2 × {10^{-6}{K^{-1}}} $$

Thermal Expansion Calculator


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Summary