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Question: A 1.0 kg bar of copper is heated at an atmospheric pressure \(P=1.01\times {{10}^{5}}N/{{m}^{2}}\). ...

A 1.0 kg bar of copper is heated at an atmospheric pressure P=1.01×105N/m2P=1.01\times {{10}^{5}}N/{{m}^{2}}. If its temperature increases from 20∘C to 50∘C20{}^\circ C\,to\,{{50}^{\circ }}C, the change in the initial energy is 2322p2322p (in J). What is the initial value of p? α=7.0×10−6/∘C\alpha =7.0\times {{10}^{-6}}/{}^\circ C, ρ=8.92×103kg/m3\rho =8.92\times {{10}^{3}}kg/{{m}^{3}}, c=387J/kg∘Cc=387J/kg{}^\circ C

Explanation

Solution

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=1.01×105N/m2P=1.01\times {{10}^{5}}N/{{m}^{2}}
Initial temperature, Ti=20∘C{{T}_{i}}=20{}^\circ C
Final temperature, Tf=50∘C{{T}_{f}}=50{}^\circ C
Thermal coefficient of the linear expansion, α=7.0×10−6/∘C\alpha =7.0\times {{10}^{-6}}/{}^\circ C
Density of copper, ρ=8.92×103kg/m3\rho =8.92\times {{10}^{3}}kg/{{m}^{3}}
Specific heat capacity of copper bar, c=387J/kg∘Cc=387J/kg{}^\circ C
Using first law of thermodynamics,
ΔQ=ΔU+W\Delta Q=\Delta U+W
Where,
ΔQ=\Delta Q=Heat absorbed/lost
ΔU=\Delta U=Change in internal energy
W=W=Work done
As we know thatΔQ=mcΔT\Delta Q=mc\Delta T
Here, ΔT\Delta Tis the change in temperature=Tf−Ti=(50−20)∘C=30∘C={{T}_{f}}-{{T}_{i}}=\left( 50-20 \right){}^\circ C=30{}^\circ C
Hence,
ΔQ=1(387)(30)J=11610J\Delta Q=1\left( 387 \right)\left( 30 \right)J=11610J
W=∫PdVW=\int{PdV}
dVdV is the change in volume=Vo(3α)ΔT={{V}_{o}}\left( 3\alpha \right)\Delta T
Hence,
W=P(mρ)(3α)ΔT =(1.01×105)(18.92×103)(3×7×10−6)(30)J =0.0071J\begin{aligned} & 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{aligned}
Putting the values ΔQ=11610J and W=0.0071J\Delta Q=11610J\text{ and }W=0.0071Jin first law of thermodynamics, we get
11610J=ΔU+0.0071J ΔU=11610J−0.0071J =11609.99J\begin{aligned} & 11610J=\Delta U+0.0071J \\\ & \Delta U=11610J-0.0071J \\\ & =11609.99J \end{aligned}
It is given that the change in internal energy is 2322p2322p
Comparing the given value of change in internal energy with the given value, we get
2322p=11609.99 p=11609.992322 p=4.999 p≈5\begin{aligned} & 2322p=11609.99 \\\ & p=\dfrac{11609.99}{2322} \\\ & p=4.999 \\\ & p\approx 5 \end{aligned}

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.