## The latent heat of fusion of water at 0 °C is 6.025 kJ mol” and the molar heat capacities (C ) of water and ice are 75.3 and 37.

Question

The latent heat of fusion of water at 0 °C is 6.025 kJ mol” and the molar heat
capacities (C ) of water and ice are 75.3 and 37.7JK’mol’, respectively. Calculate
AH for the freezing of 1 mol of supercooled water at -10.0°C.

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1 year 2021-09-03T03:52:48+00:00 1 Answers 66 views 0

$$\Delta H_{tot} = 2258.025\,kJ$$

Explanation:

The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of $$25^{\circ}C$$. Then:

$$\Delta H_{tot} = \Delta H_{s, w} + \Delta H_{f,w} + \Delta H_{s,i}$$

$$\Delta H_{tot} = n\cdot (c_{w}\cdot \Delta T_{w} + L_{f} + c_{i}\cdot \Delta T_{i})$$

Finally, the amount of heat released from water is now computed by replacing variables:

$$\Delta H_{tot} = (1\,mol)\cdot \left[\left(75.3\,\frac{kJ}{mol\cdot K} \right)\cdot (25^{\circ}C-0^{\circ}C)+ 6.025\,\frac{kJ}{mol} + \left(37.7\,\frac{kJ}{mol\cdot K} \right)\cdot (0 + 10^{\circ}C)\right]$$$$\Delta H_{tot} = 2258.025\,kJ$$