Question

Question 6.10 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. ΔfusH = 6.03 kJ mol–1 at 0°C.

Cp[H2O(l)] = 75.3 J mol–1 K–1

Cp[H2O(s)] = 36.8 J mol–1 K–1

Class XI Thermodynamics Page 183

Answer 1

conversion of 1 mole of water at 10°C to ice at -10°C involves following steps:
step 1:- enthalpy change for changing temperature from 10°to 0°C
1 mol H2O(l) at 10°C-->1 mol H2O(l) at 0°C
$H_1=C_p[H2O(l)]\times\:\triangle{T}$

= 75.3 j/mol/K × (0-10)K
= -753 j/mol= -0.753KJ/mol

step 2 :- enthalpy of fusion.
1mol H2O(l)at 0°C-->1mol H2O(s) at 0°C
$H_2=H_{freezing}$

=-6.03 KJ/mol

Step 3:- enthalpy change for changing temperature 0°C to -10°C.
1mol H2O(s) at 0°C -->1 mol H2O(s) at -10°C
$H_3=C_p[H2O(s)]\times\:\triangle{T}$

=-36.8j/mol/K×{0-(-10)}
=-368 j/mol=-0.368 KJ/mol

$then, H_{total}=H_1+H_2+H_3$

= (-0.753 -6.03 -0.368)KJ/mol
= -7.151 KJ/mol

Answer 2

Answer:

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Explanation:

conversion of 1 mole of water at 10°C to ice at -10°C involves following steps:

step 1:- enthalpy change for changing temperature from 10°to 0°C

1 mol H2O(l) at 10°C-->1 mol H2O(l) at 0°C

= 75.3 j/mol/K × (0-10)K

= -753 j/mol= -0.753KJ/mol

step 2 :- enthalpy of fusion.

1mol H2O(l)at 0°C-->1mol H2O(s) at 0°C

=-6.03 KJ/mol

Step 3:- enthalpy change for changing temperature 0°C to -10°C.

1mol H2O(s) at 0°C -->1 mol H2O(s) at -10°C

=-36.8j/mol/K×{0-(-10)}

=-368 j/mol=-0.368 KJ/mol

= (-0.753 -6.03 -0.368)KJ/mol

= -7.151 KJ/mol