Answer to Question #133186 in General Chemistry for Javu

Question

Using the values for the heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following.

kilojoules given off when 14 g of steam condenses at 100∘C and the liquid cools to 0 ∘C

Answer 1

The energy released happens in two steps;

(i) During condensation of steam to 100^oC (E_{(Phase change)})

(ii) During cooling from 100^oC to 0^oC (E_(Cooling))

Step I: Energy released during phase change depend on the mass (m), and the Latent heat of vaporization (L_v) of the water.

Thus;

E_{(Phase change)} $= mLv$

$= 14g x 2260J/g$

$= 31640J$

Step II : Energy released during cooling depend on the mass(m), specific heat capacity (c), and the temperature change ($\Delta$T)

Thus;

E_(cooling) $= mc$$\Delta$T

= 14g x 4.2J/g/K x (100-0)K

= 5880J

Step III: Total energy released is the sum of energy released in each of the steps

E_(Release) = E_{(Phase change)} + E_(Cooling)

= 31640J + 5880J

= 37520J (divide by 1000 to change to kJ)

= 37.52kJ