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Thermal Properties of Matter

JEE Main level, numerical answer, medium

73 of 126

Question 73 of 126

$M$ grams of steam at ${100}^{o}C$ is mixed with $200g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at ${40}^{o}C$ [heat of vaporization of water is $540cal/g$ and heat of fusion of ice is $80cal/g$ ], the value of $M$ is ________

Show answer and solution

Answer: 40

{M}_{ice}{L}_{f}+{\text{M}}_{ice}\left(40-0\right){C}_{w}={M}_{steam}Lv+{M}_{steam}\left(100-40\right){C}_{w}

\Rightarrow 200\left[80+40\left(1\right)\right]=M\left[540+60\left(1\right)\right]

\Rightarrow 200\left(120\right)=M\left(600\right)

M=40gm

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JEE Main 2025 Apr 2 Shift 1 question paper

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