Question

If the heat of fusion for water is 6.01 kJ/mol, the heat of vaporization is 40.79
kJ/mol, calculate the heat transferred during the deposition of 2.0 g water vapor.
Deposition is the phase change from a vapor to a solid.

Answer 1

In the process of deposition ,We have three steps.

1. Coversion of vapour to water at boiling point of water.The heat evolved in this step be Q 1.

Q1 =L_v n

L_v - Latent heat of vaporisation = 40.79 kj/mol

n - no of moles = Wt of water / gm molecular wt of water

= 2/18= 0.11 moles

Q 1 = 40.79 x 0.11 = 4.53 KJ

2. Conversion of water at 373 K to Water at 273 k. Let heat evolved in this process be Q 2.

Q 2. = n C_p ( T₁- T₂)

We know C_P Of Water is 0.0753 KJ/mol-K.

Q2 = 0.11* 0.0753* 100

= 0.8 KJ

3. Conversion of water at 273 K to ice at 273 K.let Q 3 be heat evolved in this step.

Q 3 = L_f n

L_ff - latent heat of fusion = 6.01 Kj/ mol

Q3 = 6.01 * 0.11

= 0.661 KJ

So total heat evolved in the deposition if 2 gm of water is Q = Q1 +Q2+Q3

= 4.53 +0.8+ 0.661

= 5.991 KJ