1. Heating the ice from -50°C to 0°C:
* Specific heat capacity of ice (c_ice): 2.09 J/g°C
* Mass of ice (m): 1 mol * 18.015 g/mol = 18.015 g
* Temperature change (ΔT): 0°C - (-50°C) = 50°C
* Heat absorbed (q1): q1 = m * c_ice * ΔT = 18.015 g * 2.09 J/g°C * 50°C = 1884.3 J
2. Melting the ice at 0°C:
* Enthalpy of fusion (ΔH_fus): 6.01 kJ/mol = 6010 J/mol
* Heat absorbed (q2): q2 = ΔH_fus = 6010 J
3. Heating the water from 0°C to 70°C:
* Specific heat capacity of water (c_water): 4.18 J/g°C
* Mass of water (m): 18.015 g
* Temperature change (ΔT): 70°C - 0°C = 70°C
* Heat absorbed (q3): q3 = m * c_water * ΔT = 18.015 g * 4.18 J/g°C * 70°C = 5233.7 J
Total enthalpy change (ΔH):
ΔH = q1 + q2 + q3 = 1884.3 J + 6010 J + 5233.7 J = 13128 J or 13.13 kJ
Therefore, the enthalpy change for converting 1 mol of ice at -50°C to water at 70°C is approximately 13.13 kJ.
