When liquid is heated and evaporated, some of the heating goes towards working (

Question

When liquid is heated and evaporated, some of the heating goes towards working (since the specific volume changes), while the remainder goes towards increasing the internal energy. Calculate the work (per unit mass) and the change in specific internal energy when water at 100 is transformed to water vapor. What percentage of the applied heating went towards work and internal energy changes, respectively? At 100 the density of water is 958.4 kg m-3. You may assume an external pressure of 1013 mb. When liquid is heated and evaporated, some of the heating goes towards working (since the specific volume changes), while the remainder goes towards increasing the internal energy. Calculate the work (per unit mass) and the change in specific internal energy when water at 100 is transformed to water vapor. What percentage of the applied heating went towards work and internal energy changes, respectively? At 100 the density of water is 958.4 kg m-3. You may assume an external pressure of 1013 mb.

Explanation / Answer

The heat required for conversion of 1 gram of water to vapour is = 2257 J / g

Now work done = Pressure X change in volume

Pressure = 1013mb = 1 atm

Change in volume = Volume of vapour - volume of liquid

molar volume of water vapour = RT / P = 0.0821 X 398 / 1 = 32.67 L

so volume of 1 gram = 32.67 / 18 = 1.82 L

Density of water = 958.4 Kg / m^3

So volume of 1 gram of water = mass / density = 10^-3 Kg / 958.4 Kg / m^3

= 0.00104 X 10^-3 m^3 = 0.00104 dm^3 = 0.00104 L

So change in volume = 1.82 - 0.00104 = 1.818 L

So work done = 1 X 1.818 = 184.2 Joules / g

So internal energy = 2257 J / g - 184.2 = 2072.8 Joules

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