$100$ mol$/$s of a liquid mixture containing $51$ mol $\%$ n$-$hexane and $49$ mol $\%$ h$-$pentane at $69$oC and $4\mathrm{.}6$ atm is fed to an evaporator. When the feed is exposed to the reduced pressure in the evaporator, a substantial amount is vaporized. The vapor and liquid phases, which are in equilibrium with each other, are separated and discharged as separate streams at $51\mathrm{.}5$oC$.$ The liquid product contains $42$ mol $\%$ n$-$hexane whereas the vapor product contains $9$ mol $\%$ n$-$hexane. How much heat must be transferred to the evaporator?