Question 182: Thermodynamics & KTG

Question

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 $\alpha=3.1 	imes 10^{-5} \mathrm{~K}^{-1}$ 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 $8.1 	imes 10^2 \mathrm{~J}$ then change in area of the rectangular sheet is :

Accepted Answer

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

Answer

$2.0 	imes 10^{-6} \mathrm{~m}^2$

Answer

$6.0 	imes 10^{-7} \mathrm{~m}^2$

Answer

$3.0 	imes 10^{-7} \mathrm{~m}^2$

Answer

$4.0 	imes 10^{-7} \mathrm{~m}^2$

Thermodynamics & KTG

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