Answer:
[tex]\Delta H_T=108,620J=108.6kJ[/tex]
Explanation:
Hello!
In this case, since we can evidence that the ice is firstly undergoing a melting process at constant 0.0 °C, whose associated enthalpy change is:
[tex]\Delta H_1 =m\Delta H_{fus}[/tex]
Next, the formed liquid water undergoes a heating from 0.0 °C to 22°C, to the associated enthalpy change is:
[tex]\Delta H_2 =mC_{liq}(22\°C-0.0\°C)[/tex]
Thus, the total enthalpy change, or heat added to the system turns out:
[tex]\Delta H_1 =255g*334\frac{J}{g}=85,170J\\\\ \Delta H_2=255g*4.18\frac{J}{g\°C}*(22\°C-0.0\°C)=23,449.8J\\\\ \Delta H_T=85,170J+23,449.8J\\\\\Delta H_T=108,620J=108.6kJ[/tex]
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