A rectangular metallic sheet with a linear expansivity of 1.8 \times 10^{-5}\text{ K}^{-1}… — JAMB UTME

A rectangular metallic sheet with a linear expansivity of $1.8 \times 10^{-5}\text{ K}^{-1}$ experiences a temperature rise of $50\text{ K}$ . If the increase in its surface area is $0.90\text{ cm}^2$ , what was the initial surface area of the sheet in $\text{cm}^2$ ?

Cevap: 500 cm^2

Cevap

The initial surface area of the metallic sheet is

500\text{ cm}^2

The initial area is found by converting linear expansivity to area expansivity (\beta = 2\alpha = 3.6 \times 10^{-5}\text{ K}^{-1}) and substituting into the area expansion relation \Delta A = A_0 \beta \Delta T, giving A_0 = \frac{0.90}{3.6 \times 10^{-5} \times 50} = 500\text{ cm}^2$.

Adım Adım Çözüm

Calculate the area (superficial) expansivity (\beta)

\beta = 2\alpha = 2 \times 1.8 \times 10^{-5}\text{ K}^{-1} = 3.6 \times 10^{-5}\text{ K}^{-1}

Surface area expansion depends on area expansivity, which is twice the linear expansivity for an isotropic solid.

Formulate the thermal area expansion equation

\Delta A = A_0 \beta \Delta T

The fractional change in area is directly proportional to the area expansivity and the temperature change.

Rearrange the formula to solve for the initial surface area (A_0)

A_0 = \frac{\Delta A}{\beta \Delta T}

Isolating the required unknown quantity.

Substitute the known numerical values and compute

A_0 = \frac{0.90\text{ cm}^2}{(3.6 \times 10^{-5}\text{ K}^{-1})(50\text{ K})} = \frac{0.90}{1.8 \times 10^{-3}} = 500\text{ cm}^2

Evaluating the expression yields the exact initial surface area.

Anahtar Kavram

Relationship between Linear Expansivity and Area Expansivity

Alternatif Yöntem

Calculate fractional area expansion per kelvin: \beta = 2\alpha = 3.6 \times 10^{-5}\text{ K}^{-1}

. Total fractional expansion for

50\text{ K}

is

3.6 \times 10^{-5} \times 50 = 0.0018

. Then initial area

A_0 = \frac{0.90}{0.0018} = 500\text{ cm}^2$.

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