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(CO)-1$$ and latent heat of steam $$=$$ $$540$$ cal $$g-1)

Question

Infinity Learn · Accepted Answer

Heat lost $$=$$ Heat gained⇒      $$mLv+msw$$∆θ$$=mwSw$$∆θ ⇒    m×$$540+m$$×$$1$$×(100-80)                              $$=20$$×$$1$$×(80-10) ⇒                     $$m=2.5$$ g Total mass of water $$=$$ (20$$ $$+$$ $$2.5) g $$=$$ $$22.5$$ g

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
(C^O)^-1 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(CO)-1 and latent heat of steam = 540 cal g-1)

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Ready to Test Your Skills?

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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
(C^O)^-1 and latent heat of steam = 540 cal g^-1)