Question

A 1.0 kg bar of copper is heated at an atmospheric pressure $P=1.01\times {{10}^{5}}N/{{m}^{2}}$. If its temperature increases from $20{}^\circ C\,to\,{{50}^{\circ }}C$, the change in the initial energy is $2322p$ (in J). What is the initial value of p? $\alpha =7.0\times {{10}^{-6}}/{}^\circ C$, $\rho =8.92\times {{10}^{3}}kg/{{m}^{3}}$, $c=387J/kg{}^\circ C$

Answer 1

Hint:To find the change in internal energy of the copper we use the first law of thermodynamics. To find the value of heat absorbed by the copper we use the law of calorimeters.

Complete step-by-step solution:
Given that
Mass of the copper bar 1.0 kg
Atmospheric pressure, $P=1.01\times {{10}^{5}}N/{{m}^{2}}$
Initial temperature, ${{T}_{i}}=20{}^\circ C$
Final temperature, ${{T}_{f}}=50{}^\circ C$
Thermal coefficient of the linear expansion, $\alpha =7.0\times {{10}^{-6}}/{}^\circ C$
Density of copper, $\rho =8.92\times {{10}^{3}}kg/{{m}^{3}}$
Specific heat capacity of copper bar, $c=387J/kg{}^\circ C$
Using first law of thermodynamics,
$\Delta Q=\Delta U+W$
Where,
$\Delta Q=$Heat absorbed/lost
$\Delta U=$Change in internal energy
$W=$Work done
As we know that$\Delta Q=mc\Delta T$
Here, $\Delta T$is the change in temperature$={{T}_{f}}-{{T}_{i}}=\left( 50-20 \right){}^\circ C=30{}^\circ C$
Hence,
\[\Delta Q=1\left( 387 \right)\left( 30 \right)J=11610J\]
$W=\int{PdV}$
$dV$ is the change in volume$={{V}_{o}}\left( 3\alpha \right)\Delta T$
Hence,
$\begin{align}
  & W=P\left( \dfrac{m}{\rho } \right)\left( 3\alpha \right)\Delta T \\
 & =\left( 1.01\times {{10}^{5}} \right)\left( \dfrac{1}{8.92\times {{10}^{3}}} \right)\left( 3\times 7\times {{10}^{-6}} \right)\left( 30 \right)J \\
 & =0.0071J
\end{align}$
Putting the values $\Delta Q=11610J\text{ and }W=0.0071J$in first law of thermodynamics, we get
$\begin{align}
  & 11610J=\Delta U+0.0071J \\
 & \Delta U=11610J-0.0071J \\
 & =11609.99J
\end{align}$
It is given that the change in internal energy is $2322p$
Comparing the given value of change in internal energy with the given value, we get
$\begin{align}
  & 2322p=11609.99 \\
 & p=\dfrac{11609.99}{2322} \\
 & p=4.999 \\
 & p\approx 5
\end{align}$

Hence, the value of p is 5.

Note:We assume this process as an isobaric process where pressure is constant and equal to the atmospheric pressure.
We assume that there is no heat loss to the surrounding of the copper.