Question

Consider a rectangular sheet of solid material of length $ \ell = 9 $ cm and width $ d = 4 $ cm. The coefficient of linear expansion is $ \alpha = 3.1 \times 10^{-5} $ K$^{-1}$ at room temperature and one atmospheric pressure. The mass of the sheet is $ m = 0.1 $ kg and the specific heat capacity $ C_v = 900 $ J kg$^{-1}$K$^{-1}$. If the amount of heat supplied to the material is $ 8.1 \times 10^2 $ J, then the change in area of the rectangular sheet is:

• A. \( 2.0 \times 10^{-6} \, \text{m}^2 \)
• B. \( 3.0 \times 10^{-7} \, \text{m}^2 \)
• C. \( 6.0 \times 10^{-7} \, \text{m}^2 \)
• D. \( 4.0 \times 10^{-7} \, \text{m}^2 \)

Hint:

The change in area of a material due to thermal expansion can be calculated using the formula \( \Delta A = A \alpha \Delta T \), where \( A \) is the initial area, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the temperature change.

Answer 1

The Correct Option is A

This problem addresses the calculation of the change in area of a heated rectangular sheet. The initial step is to determine the temperature change experienced by the material.

1. Determine the temperature change, denoted as \(\Delta T\):

Provided data:

• Heat supplied, \(Q = 8.1 \times 10^2 \, \text{J}\)
• Mass of the sheet, \(m = 0.1 \, \text{kg}\)
• Specific heat capacity, \(C_v = 900 \, \text{J kg}^{-1}\text{K}^{-1}\)

The heat transfer equation is:

\(Q = m \cdot C_v \cdot \Delta T\)

Rearranging this equation yields:

\(\Delta T = \frac{Q}{m \cdot C_v}\)

Substituting the provided values:

\(\Delta T = \frac{8.1 \times 10^2}{0.1 \cdot 900} = \frac{8.1 \times 10^2}{90} = 9 \, \text{K}\)

1. Calculate linear expansion and the subsequent change in area:

The area change resulting from thermal expansion is contingent on the coefficients of linear expansion.

The coefficient of linear expansion is given as \(\alpha = 3.1 \times 10^{-5} \, \text{K}^{-1}\).

For a rectangular sheet, the approximate change in area, \(\Delta A\), is expressed as:

\(\Delta A = A_0 \cdot 2 \alpha \cdot \Delta T\)

Here, \(A_0\) represents the initial area.

Calculate the initial area, \(A_0\):

\(\ A_0 = \ell \cdot d = 9 \, \text{cm} \times 4 \, \text{cm} = 36 \, \text{cm}^2 = 36 \times 10^{-4} \, \text{m}^2\)

Substitute the calculated values to determine \(\Delta A\):

\(\Delta A = 36 \times 10^{-4} \cdot 2 \cdot 3.1 \times 10^{-5} \cdot 9\)

Perform the calculation:

\(\Delta A = 36 \times 10^{-4} \cdot 6.2 \times 10^{-5} \cdot 9 = 36 \cdot 6.2 \cdot 9 \times 10^{-9}\)

\(\Delta A = 2008.8 \times 10^{-9} \, \text{m}^2 = 2.0088 \times 10^{-6} \, \text{m}^2\)

Consequently, the approximate change in area is \(2.0 \times 10^{-6} \, \text{m}^2\).

The definitive result is:

\(2.0 \times 10^{-6} \, \text{m}^2\)