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

Not all the water will vaporize because the heat absorbed by the water is less than the heat required for vaporization.

Step-by-step explanation:

To determine if all the water will vaporize, we need to calculate the amount of heat required to vaporize the water sample. First, calculate the heat absorbed by the water to raise its temperature from 27.1°C to 100°C:

1. 
   
3. q = m × Cs × ΔT
4. 
   
6. q = (35.0 g) × (4.184 J/°Cg) × (100°C - 27.1°C)
7. 
   
9. q ≈ 34,950 J

Next, calculate the heat absorbed during vaporization:

1. 
   
3. q = (35.0 g) / (18.02 g/mol) × (40.7 kJ/mol)
4. 
   
6. q ≈ 91.8 kJ

Therefore, the total heat absorbed by the water is approximately 34,950 J + 91.8 kJ = 126,750 J. Since the sample absorbed less heat energy than the heat required for vaporization, not all of the water will vaporize.

Answer 2

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* c_(p,l)* (T_(final)-T_(initial))]+m* \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=(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* 4.184J/g^oC* (100-27.1)^oC]+35.0g* 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.