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.
​

Answer 1

`\Delta H_(tot) = 2258.025\,kJ`

Step-by-step 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\,(kJ)/(mol\cdot K) \right)\cdot (25^(\circ)C-0^(\circ)C)+ 6.025\,(kJ)/(mol) + \left(37.7\,(kJ)/(mol\cdot K) \right)\cdot (0 + 10^(\circ)C)\right]`
`\Delta H_(tot) = 2258.025\,kJ`