JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 3)
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
40
Explanation
M × 540 + M + 60 = 200 × 80 + 200 × 1× (40– 0)
$$ \Rightarrow $$ 600 M = 24000
$$ \Rightarrow $$ M = 40
$$ \Rightarrow $$ 600 M = 24000
$$ \Rightarrow $$ M = 40
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