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Second law of thermodynamics, entropy

Problem 2.143

Find the entropy increment of an aluminum bar of mass m=3.0 kgm=3.0 \mathrm{~kg} on its heating from the temperature T1=300 KT_{1}=300 \mathrm{~K} up to T2=600 KT_{2}=600 \mathrm{~K} if in this temperature interval the specific heat capacity of aluminum varies as c=a+bT,c=a+b T, where a=0.77 J/(g⋅K)a=0.77 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K}) b=0.46 mJ/(g⋅K2)b=0.46 \mathrm{~mJ} /\left(\mathrm{g} \cdot \mathrm{K}^{2}\right)

Reveal Answer
 2.143. ΔS=m[aln⁡(T2/T1)+b(T2−−T1)]=2.0 kJ/K\begin{aligned} &\text { 2.143. } \Delta S=m\left[a \ln \left(T_{2} / T_{1}\right)+b\left(T_{2}-\right.\right.\\ &\left.\left.-T_{1}\right)\right]=2.0 \mathrm{~kJ} / \mathrm{K} \end{aligned}
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