$hat(H)={\displaystyle \underset{120\mathrm{.}0}{\overset{170\mathrm{.}0}{}}}[(33\mathrm{.}6{10}^{-3})+(1\mathrm{.}367{10}^{-5})T+(-1\mathrm{.}607{10}^{-8})T^2+(6\mathrm{.}473{10}^{-12})T^3]dT$

$=[(33\mathrm{.}6{10}^{-3})T+(1\mathrm{.}367{10}^{-5})(\frac{T^2}{2})+(-1\mathrm{.}607{10}^{-8})(\frac{T^3}{3})+(6\mathrm{.}473{10}^{-12})(\frac{T^4}{4})]{|}_{120\mathrm{.}0}^{170\mathrm{.}0}$

$=1\mathrm{.}76306k\frac{J}{m}ol$

Required heat input

Power $=\{\begin{array}{ccc}\text{:}[1\mathrm{.}76306kJ,& 21\mathrm{.}0\text{k}\text{m}\text{o}\text{l}\text{,}1\text{m}\text{i}\text{n}\text{,}1000\text{m}\text{o}\text{l}& ]\\ \text{m}\text{o}\text{l}\text{,}& \text{m}\text{i}\text{n}\text{,}60s,& 1\text{k}\text{m}\text{o}\text{l}\end{array}=617\mathrm{.}07kW$

eat Input for Closed Vessel

You will have to determine $hat(U),hat(U)$ in the course of the calculation.

Carry sufficient significant figures throughout the calculation.

Calculate the heat input required to raise the temperature of $21\mathrm{.}0$kmol of chlorine in a closed rigid vessel from $120\mathrm{.}0C$ to

$170\mathrm{.}0C$ initially at $1$atm.

$($Suggestion: Evaluate $hat(U)$ directly from the result of the first calculation, so that you do not have to perform another integration.$)$

Heat input: i MJ