Steam at 100°C is passed over 1000 g of ice at 0°C. After sometime 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 vaporisation of
steam (given specific heat capacity of water = 4200 J/kg°C and specific heat of fusion of
ice = 336,000 J/kg)
Answers
Answer:
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Explanation:
Let ‘L’ J/kg be the specific latent heat of vaporisation of steam.
Mass of ice melted = (1000 – 600)g = 400g = 0.4kg
∴ Mass of steam condensed into water = (450 – 400) = 50g = 0.05kg
(i) Heat given out by steam in condensing into water at the same temperature, i.e., 1000C
= mL = 0.05 x L = 0.050 L J
(ii) Heat given out by 0.050 kg of water at 1000C in cooling down to 00C
= mst = (0.50 x 4200 x 100)J = 21000 J
(iii) Heat taken up by ice in melting into water at the same temperature
= ml = 0.4 x 336000 J
Heat given out = Heat taken up
0.050 L + 21000 = 0.4 x 33600
L = 0.4 x 336000 – 21000/0.050 = 2268000 J/ kg.
Latent heat of vaporisation os steam = 2268000 J/kg
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