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

<p><strong>Calculate the Change in Temperature:</strong></p> <p>Use the formula:</p> <p>$ \Delta Q = mC_v \Delta T $</p> <p>Given:</p></p> <p><p>$\Delta Q = 8.1 \times 10^2 \, \text{J}$</p></p> <p><p>$m = 0.1 \, \text{kg}$</p></p> <p><p>$C_v = 900 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1}$</p> <p>Substitute the values to find $\Delta T$:</p> <p>$ 8.1 \times 10^2 = 0.1 \times 900 \times \Delta T $</p> <p>Solving for $\Delta T$:</p> <p>$ \Delta T = \frac{8.1 \times 10^2}{0.1 \times 900} = 9 \, \text{K} $</p></p> <p><p><strong>Calculate the Change in Area:</strong></p> <p>The change in area $\Delta A$ is calculated using:</p> <p>$ \Delta A = A_0 \times 2 \alpha \Delta T $</p> <p>Where:</p></p> <p><p>$A_0$ is the original area of the sheet.</p></p> <p><p>$\alpha = 3.1 \times 10^{-5} \, \text{K}^{-1}$.</p></p> <p><p>$\Delta T = 9 \, \text{K}$.</p> <p>First, calculate the original area $A_0$:</p> <p>$ A_0 = l \times d = 9 \, \text{cm} \times 4 \, \text{cm} = 36 \, \text{cm}^2 = 0.0036 \, \text{m}^2 $</p> <p>Now, substitute into the formula for $\Delta A$:</p> <p>$ \Delta A = 0.0036 \times 2 \times 3.1 \times 10^{-5} \times 9 $</p> <p>Simplifying:</p> <p>$ \Delta A = 2.0 \times 10^{-6} \, \text{m}^2 $</p></p> <p>Therefore, the change in area of the rectangular sheet is $\boxed{2.0 \times 10^{-6} \, \text{m}^2}$.</p>