Question

Calculate the heat energy absorbed or released to evaporate 6g of liquid water.

Answer 1

`Q_(net) = 15469.5\,J`

Step-by-step explanation:

Let assume that liquid water has an initial temperature of 25°C and heating and evaporation process occurs under a pressure of 1 atmosphere. Then, the heat absorbed is equal to the sum of sensible and latent components:

`Q_(net) = Q_(s) + Q_(l)`

`Q_(net) = (6\,g)\cdot \left[\left(4.186\,(J)/(g\cdot ^(\circ)C)\right)\cdot (100\,^(\circ)C-25\,^(\circ)C)+2264.3\,(J)/(g) \right]`

`Q_(net) = 15469.5\,J`