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 40 °C [heat of vaporisation of water is 540 cal g⁻¹ and heat of fusion of ice is 80 cal g⁻¹], the value of M is .

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Accepted Answer

40. Given: everything ends as liquid water at 40 °C, so the steam condenses fully and the ice melts fully. Heat given out by the steam: it condenses, then the condensate cools from 100 °C to 40 °C, so Qₒᵤₜ=M×540+M×1×60=600M cal. Heat taken in by the ice: it melts, then the melt-water warms from 0 °C to 40 °C, so Qᵢₙ=200×80+200×1×40=24000 cal. 600M=24000, so M=40.

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 40 °C [heat of vaporisation of water is 540 cal g⁻¹ and heat of fusion of ice is 80 cal g⁻¹], the value of M is ______.

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