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 $g^{-1}$ °$C^{-1}$ and latent heat of steam = 540 cal $g^{-1}$]

Question

Dr. Abubacker · Accepted Answer

[Solution] m(540 + 20) = 20 × 70 ⇒ m = 2.5 g of steam condenses; total water = 22.5 g. Answer (4).

Q. 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 $g^{-1}$ °$C^{-1}$ and latent heat of steam = 540 cal $g^{-1}$]

Explanation

Reviewed by Dr. Abubacker

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