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Thermal Physics

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3 Dec 2011 13:47:00 IST

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1 g of ice is mixed with 1 g of steam. After thermal equilibrium is achieved, temperatur

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1 g of ice is mixed with 1 g of steam. After thermal equilibrium is achieved, temperatur

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edison

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Joined: 19 Oct 2006

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5 Dec 2011 14:35:40 IST

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At thermal equilibrium the temperature of the mixture will be the same.

Let us assume that the ice is initially at 0^oC and the steam is initially at 100^oC.

The steam will give the ice some heat. And the ice will gain this heat from the steam. This will happen until the temperature of the mixture is the same.

Heat given by steam = Latent heat of vaporization of steam + Heat lost by water at 100^oC in reaching an equilibrium temperature T.

Heat gained by ice = Latent heat of melting of ice + Heat gained by water at 0^oC in reaching an equilibrium temperature T.

Now latent heat of vaporization of steam is 2258J/g at 1 bar pressure and 373K.
But latent heat of melting of ice is 334J/g at 1 bar pressure and 373 K.

Now heat gained by ice = heat lost by steam.

But we see something interesting here. We see that even if the ice melts and reaches a temperature of 100^oC, the net heat it will gain will be less than that provided by the condensation of steam!

Heat gained by ice when being heated to 100^oC is 334 + 4.2(373-273) = 754 J which is much less than the 2258 J that condensation of steam gives.

So all the ice will get converted to water at 100^oC when the equilibrium is attained and 100^oC will be the equilibrium temperature.

At equilibrium there will 1 g of water because of the ice at 100^oC.

The heat provided by the initial steam to this ice will be 754 J. Mass of steam condensed to provide this heat = 754/2258 = 0.33g.

Hence at equilibrium mass of steam at 100^oC = 0.67g and mass of water at 100^oC = 1.33 g. The equilibrium temperature is 100^oC.

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