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QJEEM20Ph5e1c: JEE Main 2020 Physics - Class 11 JEEM20Ph5e1c Solution: M grams of steam at 100 o C is mixed with 200 g of ice at its…

M grams of steam at 100oC is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produced liquid water at 40oC [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g] the value of M is ____

Step-by-Step Verified Solution

M × 540 + M + 60 = 200 × 80 + 200 × 1× (40– 0)

⇒ 600 M = 24000

⇒ M = 40

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