A 1 kg block of copper at 350°C is quenched in a 10 kg oil bath initially at ambient temperature of 20°C. Calculate the final uniform temperature (no heat transfer to/from ambient) and the change of exergy of the system (copper and oil).

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Accepted Answer

C.V. Copper and oil. [latex]C _{ co }=0.42 kJ / kg K , \quad C _{ oil }=1.8 kJ / kg K[/latex]

[latex]\begin{aligned}& m _{2} u _{2}- m _{1} u _{1}={ }_{1} Q _{2}-{ }_{1} W _{2}=0= m _{ co } C _{ co }\left( T _{2}- T _{1} ight)_{ co }+( mC )_{ oil }\left( T _{2}- T _{1} ight)_{ oil } \&1 imes 0.42\left( T _{2}-350 ight)+10 imes 1.8\left( T _{2}-20 ight)=0 \&18.42 T _{2}=507 \quad \Rightarrow \quad T =27.5^{\circ} C =300.65 K\end{aligned}[/latex]

For each mass copper and oil, we neglect work term (v = C) so Eq.8.28 is

[latex]\left(\phi_{2}-\phi_{1} ight)= u _{2}- u _{1}- T _{o}\left( s _{2}- s _{1} ight)= mC \left[\left( T _{2}- T _{1} ight)- T _{o} \ln \left( T _{2} / T _{1} ight) ight][/latex]

[latex]\begin{aligned}& m _{ cv }\left(\phi_{2}-\phi_{1} ight)_{ cv }+ m _{ ext {oil }}\left(\phi_{2}-\phi_{1} ight)_{ ext {oil }}= \&\qquad \begin{aligned}=& 0.42 imes\left[(-322.5)-293.15 \ln \frac{300.65}{623.15} ight]+10 imes 1.8\left[7.5-293.15 \ln \frac{300.65}{293.15} ight] \&=-45.713+1.698=- 4 4 . 0 k J\end{aligned}\end{aligned}[/latex]

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Eq.8.28 : [latex]\phi_{2}-\phi_{1}=\left(e_{2}-T_{0} s_{2}+P_{0} v_{2} ight)-\left(e_{1}-T_{0} s_{1}+P_{0} v_{1} ight)[/latex]

Question 8.75: A 1 kg block of copper at 350°C is quenched in a 10 kg oil b...

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