Question

Calculate the enthalpy change on freezing of $1.0\ mol$ of water at $10.0°C$ to ice at $-10.0°C$. $∆_{fus}H = 6.03\ kJ mol^{–1}$ at $0°C$.
$C_p [H_2O(l)] = 75.3\ J mol^{–1}K^{–1}$
$C_p [H_2O(s)] = 36.8 \ J mol^{–1}K^{–1}$

Answer 1

Total enthalpy change involved in the transformation is the sum of the following changes:
(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.
(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.
(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.
$Total\ △H = C_p[H_2OCl]△T+△H_{freezing}+C_p[H_2O(s)]△T$
= (75.3 J mol–1 K–1) (0 – 10)K + (–6.03 × 103 J mol–1) + (36.8 J mol–1 K–1) (–10 – 0)K
= –753 J mol–1 – 6030 J mol–1 – 368 J mol–1
= –7151 J mol–1
= –7.151 kJ mol–1

Hence, the enthalpy change involved in the transformation is –7.151 kJ mol–1.