When x grams of steam at 10 0 ∘ C is mixed with y grams of ice at 0 ∘ C, we obtain (x+y) grams of water at 10 0 ∘ C. What is the ratio y/x? ( L f of ice = 80 cal/gm , L V of steam = 540 cal/gm )

Question

When x grams of steam at 10 0 ∘ C is mixed with y grams of ice at 0 ∘ C, we obtain (x+y) grams of water at 10 0 ∘ C. What is the ratio y/x? ( L f ​ of ice = 80 cal/gm , L V ​ of steam = 540 cal/gm )

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Ans (C)

Heat given out when x grams of steam at $10 0^{\circ}$ C is converted to water at $10 0^{\circ}$ C : $H_{1} = 540 x$ ...(1)

Heat gained by ice at $0^{\circ}$ C converted to water at $10 0^{\circ}$ C : $H_{2} = 80 y + y \times 1 \times (100 - 0) = 180 y$ ...(2)

According to principle of calorimetry, $H_{1} = H_{2}$

$540 x = 180 y \Rightarrow \frac{y}{x} = 3$

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