$1 k g$ of ice at $0^{\circ} C$ is mixed with $1 k g$ of steam at $100^{\circ} C$ . What will be the composition of the system when thermal equilibrium is reached? Latent of fusion $3.36 \times 10 J / g$ and latent heat of vaporization of water = $2.26 \times 10 J / g$ .

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Solution

Heat needed to melt the ice = mL = $1 \times 3.36 \times 100000 = 336000$ J

heat needed to increase the temperature of water to 100 degree centigrade= $1 \times 4200 = 4200$ J

total heat requirement $= 340200 J$

suppose m mass of steam is converted in to water

then $m \times 2260000 = 340200 m$
$= \frac{340200}{2260000} = 150.5$ g

final composition water is $= 1150.5 g$ & steam $= 849.5 g$

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Q. 1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 10³ J kg⁻¹ and latent heat of vaporization of water = 2.26 × 10⁶ J kg⁻¹.

1 kg of ice at $0^{\circ} C$ is mixed with 1 kg of steam at $100^{\circ} C$ . What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = $3.36 \times 10^{5} J k g^{- 1}$ and latent heat of vaporization of water = $2.26 \times 10^{6} J k g^{- 1}$