Steam at 100^0C is passed into 20g of water at 10^0C When water acquires a temperature of 80^0C, the mass of water present will be: [ Take specific heat of water = 1 cal g^–1 ^0C^–1 and latent heat o…

Question

Steam at 100^0C is passed into 20g of water at 10^0C When water acquires a temperature of 80^0C, the mass of water present will be: [ Take specific heat of water = 1 cal g^–1 ^0C^–1 and latent heat of steam = 540 cal g^–1]

Accepted Answer

Correct answer: D. (d) m(g) steam at 100° → m(g) water at 100°C + 540m ......(a) m(g) water at 100°C → m(g) water at 80°C + (m)(a) (20) ......(b) (a) + (b) m(g) steam at 100°C → m(g) water at 80° + 560m (cal) ......(c) 20 g water at 10°C + (20) (a) 70 → 20 g water at 80°C ......(d) from (c) and (d) mix + 1400 cal → (20 + m) g water at 80°C + 560m (cal) 1400 = 560m 2.5 = m Total mass of water present = (20 + 2.5)g = 22.5g

$Steam at 10 0^{0} C is passed into 20g of water at 1 0^{0} C When water acquires a temperature of 8 0^{0} C, the mass of water present will be: [ Take specific heat of water = 1 cal g^{-1}^{0} C^{-1} and latent heat of steam = 540 cal g^{-1}]$

Solution

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