$M$ grams of steam at $100^{\circ} C$ is mixed with $200 g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} C$ [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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6 months ago

$M_{ice} L_{f} + m_{ice} (40 - 0) C_{w} = m_{steam} L_{v} + m_{steam} (100 - 40) C_{w}$

$\Rightarrow 200 [80 + 40 (1)] = M [540 + 60 (1)]$

$\Rightarrow 200 (120) = M (600)$

$M = 40 g m$

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