Steam at $$100$$°C is passed into $$20$$ g of water at $$10$$°C. When water acquires a temperature of $$80$$°C, the mass of water present will be [Take specific heat of water $$=$$ $$1$$ cal $$g-1oC$$ and latent heat of steam $$=$$ $$540$$ cal $$g-1$$ ]

Question

Steam at $$100$$°C  is passed into $$20$$ g of water at $$10$$°C. When water acquires a temperature of $$80$$°C, the mass of water present will be [Take specific heat of water $$=$$ $$1$$ cal $$g-1oC$$ and latent heat of steam $$=$$ $$540$$ cal $$g-1$$ ]

Infinity Learn · Accepted Answer

Here,  Specific heat of water, $$sw=1$$ cal g−$$1$$∘C−$$1$$ Latent heat of steam, $$Ls=540$$ cal g−$$1$$ Heat lost by mg of steam at $$100$$∘C to change into water  at $$80$$∘C is $$Q1=mLs+msw$$Δ$$Tw=m$$×$$540+m$$×$$1$$×$$(100$$−$$80)=540m+20m=560m$$ Heat gained by $$20g$$ of water to change its temperature  from $$10$$∘C to $$80$$∘C is $$Q2=mwsw$$Δ$$Tw=20$$×$$1$$×$$(80$$−$$10)=1400$$ According to principle of calorimetry  $$Q1=Q2$$∴ $$560m=1400$$ or  $$m=2.5gTotal$$ mass of water present $$=(20+m)g=(20+2.5)g=22.5g$$

Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be [Take specific heat of water = 1 cal $g^{- 1 o} C$ and latent heat of steam = 540 cal $g^{- 1}$ ]

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Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be [Take specific heat of water = 1 cal g-1oC and latent heat of steam = 540 cal g-1 ]

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Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be [Take specific heat of water = 1 cal $g^{- 1 o} C$ and latent heat of steam = 540 cal $g^{- 1}$ ]