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 ____
Answer (integer)
40
Solution
M × 540 + M + 60 = 200 × 80 + 200 × 1× (40– 0)
<br><br>$\Rightarrow$ 600 M = 24000
<br><br>$\Rightarrow$ M = 40
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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