Answer to Question #129747 in Molecular Physics | Thermodynamics for Mubina

Question

Suppose molar specific heat C_V of iron is given by C_V=AT^5, where A is some constant. For iron, A=5.17×〖10〗^(-3) J⁄(mol .K^4 ). Calculate the entropy change for 2.00 mol of iron when its temperature increases from 15 K to 25 K.

Answer 1

By the definition, $C_V=\frac{dQ}{dT}$, where Q is heat, transferred to the 1 mole of material, T - its temperature. Hence, $dQ=C_VdT$. The enthropy change $dS=\frac{dQ}{T}$. Thus, for n moles: $dS=\frac{n\cdot C_VdT}{T}$. If $C_V=AT^5$:

$dS=\frac{n\cdot AT^5dT}{T}=n\cdot AT^4dT \Rightarrow ΔS=S_2-S_1=\frac{n\cdot A }{5}\cdot (T^5_2-T^5_1)=$

$=\frac {2\cdot 5.17 \cdot 10^{-3}}{5} \cdot (9.77 \cdot 10^{6}-0.76\cdot 10^{6})=18.6 \cdot 10^3 \frac {J}{K}$