Question

A 35.0 g sample of water at 27.1°C absorbs 104.2 kJ of heat energy. Will all the water vaporize? Assume the water is placed in an insulated and closed container. (molar mass = 18.02 g/mol, Csliquid = 4.184 J/°Cg, Csgas = 2.01 J/°Cg, ΔHvap = 40.7 kJ/mol)

Answer 1

Answer : Yes, all the water will vaporize.

Solution :

We have to determine the total heat absorbed by the sample.

The process involved in this problem are :

$(1):H_2O(l)(27.1^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}$

where,

m = mass of sample = 35.0 g

$c_{p,l}$ = specific heat of liquid water = $4.184J/g^oC$

$\Delta H_{vap}$ = enthalpy change for fusion  = $40.7kJ/mole=40700J/mole=\frac{40700J/mole}{18.02g/mole}J/g=2258.6J/g$

Now put all the given values in the above expression, we get:

$Q=[35.0g\times 4.184J/g^oC\times (100-27.1)^oC]+35.0g\times 2258.6J/g$

$Q=89726.476J=89.73kJ$

From this we conclude that the calculated heat energy is less than the given heat energy that means all the water will vaporize.

Hence, all the water will vaporize.