The molar heat capacity of a certain substance varies with temperature according to the empirical equation

$\mathbf{C}\mathbf{=}\mathbf{}\mathbf{29}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{mol}\mathbf{\times }\mathbf{K}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{mol}\mathbf{\times }{\mathbf{K}}^{\mathbf{2}}\mathbf{)}\mathbf{T}$How much heat is necessary to change the temperature of 3.00 mol of this substance from $\mathbf{27}\mathbf{\hspace{0.33em}}\mathbf{º}\mathbf{C}$to $\mathbf{227}\mathbf{\hspace{0.33em}}\mathbf{º}\mathbf{C}$? (Hint: Use Eq. (17.18) in the form dQ = nCdT and integrate.)

The amount of heat energy that needs to be added to one mole of a substance so that there is a unit increase in its temperature is known as the molar heat capacity. Mathematically,

${\mathbf{c}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{C}}{\mathbf{n}}$

Where, n is the number of moles, C is the heat capacity and cm is the molar heat capacity.

The differential equation to be used is,

$dQ=nCdT$ ..(i)

The specific heat expression is given as,

$C\left(T\right)=a+bT$

Integrate equation (i) with respect to the temperature and setting the limits at T1and T2

$$\begin{gathered} Q=\mathrm{n}{\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}\mathrm{C}\left(\mathrm{T}\right)\mathrm{dT} \\ =\mathrm{n}\left[\mathrm{a}{\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}\mathrm{dT}+\mathrm{b}{\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}\mathrm{TdT}\right] \\ =\mathrm{n}\left[\mathrm{a}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)+\frac{\mathrm{b}}{2}\left({{\mathrm{T}}^2}_2-{{\mathrm{T}}^2}_1\right)\right] \end{gathered}$$

Substitute ${\mathrm{T}}_1=27º\mathrm{C}+273=300\mathrm{K},\mathrm{T\_}2=227º\mathrm{C}+273=500\mathrm{K}$and all the values given in above equation,

$$\begin{gathered} Q=\left(3.00\mathrm{mol}\right)\left[\left(29.5\mathrm{J}/\mathrm{mol}.\mathrm{K}\right)\left(500\mathrm{K}-300\mathrm{K}\right)+\left(4*{10}^{-3}\mathrm{J}/\mathrm{mol}.{\mathrm{K}}^2\right)-{\left(300\mathrm{K}\right)}^2\right] \\ =1.97\times {10}^4\mathrm{J} \end{gathered}$$

Thus the heat required for increasing the temperature is $1.97\times {10}^4\mathrm{J}$.