Sheet $2$: Evaporation

A single$-$effect evaporator is used to concentrate $7k\frac{g}{s}$ of a solution from $10$ to $50$ percent of solids. Steam is available at $205k\frac{N}{m^2}$ and evaporation takes place at $13\mathrm{.}5k\frac{N}{m^2}.$ If the overall heat transfer coefficient is $3k\frac{W}{m^2}.K,$ calculate the heating surface required and the amount of steam used if the feed to the evaporator is at $294K$ and the condensate leaves the heating space at $352\mathrm{.}7K.$ The specific heat capacity of a $10$ percent solution is $3\mathrm{.}76k\frac{J}{k}g.$ K$,$ the specific heat capacity of a $50$ percent solution is $3\mathrm{.}14k\frac{J}{k}g.$ K$.$