Question

1 m^3 of water is converted into 1671 m^3 of steam at atmospheric pressure and 100°C temperature. the latent heat of vaporisation of water is 2.3 * 10^-6 j/kg. If 2 kg of water be converted into steam at atmospheric pressure and 100°C temperature, then how much will be the increase in its internal energy. Take density of water os 1.0 * 10^3 kg/m^3 and atmospheric pressure is 1.01 * 10^5 N/m^2

Answer 1

Answer:

Explanation:

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Here is the answer to your question  

Change in internal energy = Heat in - Work done  

This is a constant pressure situation so Work done = P(V2 - V1)  

P = 1atm = 101.3 kPa  

V2 = 1.671 x 10^-3 m^3  

V1 = 1 x 10^-6 m^3  

Work = 101300(1.67 x 10^-3) = 169.17 J  

Heat in = mass x heat of vaporization = 0.001(2.26 x 10^6) = 2260J  

Change in internal energy = 2260 - 169.17 = 2090.83 J