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

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Heat & Thermodynamics

M grams of steam at

10 0^{\circ} C

is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at

4 0^{\circ} C

[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

Solution

The heat the steam gives out as it condenses and then cools to

4 0^{\circ}

C must balance the heat the ice takes in melting and warming its meltwater to

4 0^{\circ}

C. The steam releases

M \times 540

cal on condensing plus

M \times 1 \times 60

cal on cooling, a total of

600 M

cal. The ice absorbs

200 \times 80 = 16000

cal to melt and

200 \times 1 \times 40 = 8000

cal to warm, totalling

24000

cal. Setting

600 M = 24000

gives

M = 40

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Integer answer

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