In an industrial process $$10$$ kg of water per hour is to be heated from $$20oC$$ to $$80oC$$. To do this steam at $$150oC$$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at $$90oC$$. How many kg of steam is required per hour?(Specific$$ heat of steam $$=$$ $$1$$ calorie per gmoC, Latent heat of vaporisation $$=$$ $$540$$ $$cal/gm)

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Suppose m kg steam required per hourHeat released by steam in following three steps (i) When $$150$$∘C steam ⟶$$Q1100$$∘C steam       $$Q1=mcSteam$$ Δθ$$=m$$×$$1(150$$−$$100)=50m$$ cal  (ii) When $$100$$∘C steam ⟶$$Q2100$$∘C water        $$Q2=mLV=m$$×$$540=540m$$ cal  (iii) When $$100$$∘C water ⟶$$Q390$$∘C water        $$Q3=mcW$$Δθ$$=m$$×$$1$$×$$(100$$−$$90)=10mcalHence$$ total heat given by the steam $$Q=Q1+Q2+Q3$$ $$=$$ $$600$$ mcal  ……..$$(i)Heat$$ taken by $$10$$ kg water   Q′$$=mcW$$Δθ$$=10$$×$$103$$×$$1$$×$$(80$$−$$20)=600$$×$$103cal$$ Hence $$Q=Q$$′⇒$$600m=600$$×$$103$$      ⇒$$m=103gm=1kg$$

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