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 is_______

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answer is 0040.00.

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Detailed Solution

Heat gain by ice = Heat loss by steamMiceLf+Mice40−0Cw=msteamLV+msteam100−40Cw⇒20080+401=M540+601⇒200120=M600 ⇒M=40g

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