Calculate the amount of ice that can be melted when one kg of steam at 100 degree Centigrade condenses to water at 100 Degree Centegrade. Latent heat of fusion = 33610^3 J kg^-1 and Latent heat of vapourisation = 226010^3 J Kg^-1.

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Solution

the amount of ice melted is amount of heat released by vapour while coming to water
Q₁₌mL= 1x2260000 joule
let mass of ice melted m
Q₂₌mx336000
hence heat gain = heat lost
$Q_{1} = Q_{2} 2260000 = m \times 336000 m = \frac{2260000}{336000} = 6.72 k g$

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