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Q.

M gram 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 40ºC [heat of vaporization of water is 540 cal/g and heat and heat of fusion of ice is 80 cal/ g], the value of M(in g) is_______

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answer is 40.

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Detailed Solution

$\begin{array}{l} H e a t g a i n b y i c e = H e a t l o s s b y s t e a m \\ M_{i c e} L_{f} + M_{i c e} (40 - 0) C_{w} = m_{s t e a m} L_{V} + m_{s t e a m} (100 - 40) C_{w} \\ \Rightarrow 200 [80 + 40 (1)] = M [540 + 60 (1)] \\ \Rightarrow 200 (120) = M (600) \\ \Rightarrow M = 40 g \end{array}$

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