M grams of steam at 100°C is mixed with 200g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is______.
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(40)
Sol. M ice L f + m ice (40 – 0) C w = m steam L v + m steam (100 – 40) C w
⇒ 200[80 + 40(a)] = m [540 + 60(a)]
⇒ 200 (120) = m (600)
m = 40 gm
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