Steam at $$1000C$$ is passed into $$20$$ g of water at $$100C$$. When water acquires a temperature of $$800C$$, the mass of water present will be (Take$$, specific heat of water $$=$$ $$1$$ $$calg-1$$ $$0C-1$$ and latent heat of steam $$=$$ $$540$$ cal $$g-1)

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Infinity Learn · Accepted Answer

Heat lost by system $$=$$ Heat gained by water Let  m' be amount of heat converted into waterm'×$$L=mc$$Δtm'×$$540=20$$×$$1$$×(80$$−$$10)⇒ m'$$=20$$×$$70540=2.5gNow$$, net mass of water $$=$$ $$20$$ $$+$$ $$2.5$$ $$=$$ $$22.5$$ g

Steam at 100⁰C is passed into 20 g of water at 10⁰C. When water acquires a temperature of 80⁰C, the mass of water present will be (Take, specific heat of water = 1 calg^-1 ⁰C^-1 and latent heat of steam = 540 cal g^-1)

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Steam at 1000C is passed into 20 g of water at 100C. When water acquires a temperature of 800C, the mass of water present will be (Take, specific heat of water = 1 calg-1 0C-1 and latent heat of steam = 540 cal g-1)

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Steam at 100⁰C is passed into 20 g of water at 10⁰C. When water acquires a temperature of 80⁰C, the mass of water present will be (Take, specific heat of water = 1 calg^-1 ⁰C^-1 and latent heat of steam = 540 cal g^-1)