If there are no heat losses, the heat released by the condensation of x g of steam at 1000C into water at 1000C can be used to convert y g of ice at 00C into water at 1000C. Then the ratio y:x is nearly

Heat released to convert x gm of steam at 1000C to water at 1000C is x×540 cals.If y gm of ice is converted from 00C to water at 1000C it requires heat isy×80+y×1×100=180 y∴x×540=180y ⇒ yx=540180=31

If there are no heat losses, the heat released by the condensation of x g of steam at 1000C into water at 1000C can be used to convert y g of ice at 00C into water at 1000C. Then the ratio y:x is nearly

Heat released to convert x gm of steam at 1000C to water at 1000C is x×540 cals.If y gm of ice is converted from 00C to water at 1000C it requires heat isy×80+y×1×100=180 y∴x×540=180y ⇒ yx=540180=31

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If there are no heat losses, the heat released by the condensation of x g of steam at 100⁰C into water at 100⁰C can be used to convert y g of ice at 0⁰C into water at 100⁰C. Then the ratio y:x is nearly

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$1 : 1$

$2.5 : 1$

$2 : 1$

$3 : 1$

answer is D.

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Detailed Solution

Heat released to convert x gm of steam at 100⁰C to water at 100⁰C is $x \times 540$ cals.
If y gm of ice is converted from 0⁰C to water at 100⁰C it requires heat is
$y \times 80 + y \times 1 \times 100 = 180 y$
$∴ x \times 540 = 180 y ⇒ \frac{y}{x} = \frac{540}{180} = \frac{3}{1}$