Evaporation of water

Problem

How much energy is needed to evaporate $1{,}0\;\text{kg}$ of water at $100\;^\circ\text{C}$? Use the latent heat of vaporisation $\lambda_\text{vap} = 2{,}26\cdot 10^6\;\text{J/kg}$.

Solution

Setup. The water is already at boiling temperature: there is no need to raise its temperature, only to carry out the change of state liquid → vapour. The energy required is the latent heat: $Q = m\,\lambda_\text{vap}$

Numerical calculation. $Q = 1{,}0 \cdot 2{,}26\cdot 10^6 = 2{,}26\cdot 10^6\;\text{J}$

$\ev{Q \approx 2{,}26\cdot 10^6\;\text{J}}$

During evaporation the temperature stays constant at $100\;^\circ\text{C}$: all the energy supplied goes into breaking the bonds holding the molecules together in the liquid phase, not into increasing the average kinetic energy.

Links

Topics: Thermodynamics Concepts: Latent heat and changes of state