please answer both of the questions: -If 1.00 kg of ice at 273 K comes to equili

Question

please answer both of the questions:

-If 1.00 kg of ice at 273 K comes to equilibrium with 1 kg of steam at 373 K. What is the final state of this combination at thermal equilibrium?
The specific heat, cW, of water is 4190 J/kg·K, the specific heat, cI, of ice is 2100 J/kg·K. The heat of fusion, Lf, for water is 334 x 103 J/kg, the heat of vaporization of water, Lv, is 2256 x 103 J/kg.

-A solid concrete wall 3 m by 2.1 m and 27 cm thick, with a thermal conductivity of 1.27 W/m·K, separates a basement at 18 °C from the ground outside at 6 °C. Under steady state conditions, how much heat flows through the wall per hour?

Explanation / Answer

At thermal equilibrium the temperature of the mixture will be the same , let it be T Kelvin.

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

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

Now heat gained by ice = heat lost by steam.

Heat gained by ice when being heated to 100oC is 334 * 10^3 + 4190 * (373-273) = 753 *10^3J/Kg which is much less than the 2256 * 10^3 J/Kg that condensation of steam gives.

So all the ice will get converted to water at 100oC when the equilibrium is attained and 100oC will be the equilibrium temperature.
At equilibrium there will 1 Kg of water because of the ice at 373 K.

The heat provided by the initial steam to this ice will be 753 * 10^3 J. Mass of steam condensed to provide this heat = 753 * 10^3/2256 * 10^3 = 0.33 Kg.

Hence at equilibrium mass of steam at 373oK = 0.67Kg and mass of water at 373oK  = 1.33 Kg. The equilibrium temperature is 373oK .


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