$50 g m$ ice at - $10^{0} C$ is mixed with $20 g m$ steam at $100^{0} C .$ When the mixture finally reaches its steady state inside a calorimeter of water equivalent $1.5 g m$ then : [Assume calorimeter was initially at $0^{0} C,$ Take latent heat of vaporization of water $= 540$ cal/gm, Latent heat of fusion of water $= 80$ cal/gm, specific heat capacity of water $= 1 c a l / g m -^{0} C,$ specific heat capacity of ice $= {0.5 cal/ gm}^{0} C$ ]

Mass of water remaining is :

67.4 g m

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Mass of steam remaining is :

2.6 g m

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Mass of water remaining is :

67.87 g m

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Mass of steam remaining is :

2.13 g m

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Solution

The correct options are
B Mass of steam remaining is : $2.6 g m$
C Mass of water remaining is : $67.87 g m$

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Determine the final result when 200 gm of water and 20 gm of ice at $0^{\circ} C$ are in a calorimeter having a water equivalent of 30 gm. 50 gm of steam is passed into it at $100^{\circ} C$ . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm

Determine the final result when 200 gm of water and 20 gm at $0^{\circ} C$ are in a calorimeter having a water equivalent of 30 gm and 50 gm of steam is passed into it at $100^{\circ} C$ . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm