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.

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

Answers ( )

    0
    2021-09-03T03:54:26+00:00

    Answer:

    [tex]\Delta H_{tot} = 2258.025\,kJ[/tex]

    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 [tex]25^{\circ}C[/tex]. Then:

    [tex]\Delta H_{tot} = \Delta H_{s, w} + \Delta H_{f,w} + \Delta H_{s,i}[/tex]

    [tex]\Delta H_{tot} = n\cdot (c_{w}\cdot \Delta T_{w} + L_{f} + c_{i}\cdot \Delta T_{i})[/tex]

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

    [tex]\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][/tex][tex]\Delta H_{tot} = 2258.025\,kJ[/tex]

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )