Question
$$\Delta S=\int_{T_{1}}^{T_{2}} \frac{C d T}{T}=\int_{T_{1}}^{T_{2}} \frac{m(a+b T)}{T} d T=m b\left(T_{2}-T_{1}\right)+m a \ln \frac{T_{2}}{T_{1}}$$
Step 1
0$ kg, and it is heated from temperature $T_1 = 300$ K to $T_2 = 600$ K. The specific heat capacity, $C$, varies with temperature as $C = a + bT$, where $a = 0.77$ J/g K and $b = 0.46 \times 10^{-3}$ J/g K$^2$. Show more…
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