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Question
Steam at 100°C is passed over 1000 g of ice at 0°C. After some time, 600 g of ice at 0°C is left and 450 g of water at 0°C is formed. Calculate the specific latent heat of vaporization of steam (Given: specific heat capacity of water = 4200 J/kg°C, specific latent heat of fusion of ice = 336,000 J/kg.)
Solution
Let 'L' J/kg be the specific latent heat of vaporisation of steam.
Mass of ice method = (1000 - 600)g = 400g = 0.4 kg
∴ Mass of steam condensed into water = (450 - 400) = 50g = 0.05 kg
(i) Heat given out by steam in condensing into water at the same temperature i.e., 100°C
= mL = 0.05 × L = 0.050 L J
(ii) Heat given out by 0.050 kg of water at 100°C in cooling down to 0°C
= mst = (0.50 × 4200 × 100) J = 21000 J
(iii) Heat taken up by ice in melting into water at the same temperature
= ml = 0.4 × 336000 J
Heat given out = Heat taken up
0.050 L + 21000 = 0.4 × 33600
L = `(0.4 xx 336000 - 21000)/0.050` = 2268000 J/kg
Latent heat of vaporisation of steam = 2268000 J/kg.
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