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Calorimetry Class 11th

M grams of steam at 100 ∘ C is mixed with 200 g 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 c a l / g and heat of fusion of ice is 80 c a l / g ], the value of M  is                    
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alok kumar singh

Contributor-Level 10

M ice   L f + m ice   ( 40 - 0 ) C w = m steam   L v + m steam   ( 100 - 40 ) C w

⇒ 200 [ 80 + 40 ( 1 ) ] = M [ 540 + 60 ( 1 ) ]

⇒ 200 ( 120 ) = M ( 600 )

M = 40 g m

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