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Worked Examples · Example 4.2

Q.1.0 kg of liquid water is boiled at 100 °C and all of it is converted to steam. If the change of state takes place at the atmospheric pressure (1.01 × 10⁵ Pa), calculate

(a) the energy transferred to the system,
(b) the work done by the system during this change, and
(c) the change in the internal energy of the system. (Volume changes from 1.0 × 10⁻³ m³ liquid to 1.671 m³ steam; L = 2256 kJ/kg.)
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Q = 2256 kJ absorbed, W = pΔV = 169 kJ done by the steam, so ΔU = 2256 − 169 = 2087 kJ.

  1. Energy transferred. Water becomes steam by absorbing the latent heat of vaporization:

    Q=L m=(2256 kJ/kg)(1.0 kg)=2256 kJQ = L\,m = (2256\ \mathrm{kJ/kg})(1.0\ \mathrm{kg}) = 2256\ \mathrm{kJ}

  2. Work done. At constant pressure W = pΔV with ΔV = (1.671 − 1.0×10⁻³) m³:

    W=p ΔV=(1.01×105)(1.671−1.0×10−3)=1.69×105 J=169 kJW = p\,\Delta V = (1.01\times10^{5})(1.671 - 1.0\times10^{-3}) = 1.69\times10^{5}\ \mathrm{J} = 169\ \mathrm{kJ}

  3. Change in internal energy. From the first law, …

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