Respuesta :
Answer : The enthalpy change for converting 1 mole of ice at [tex]-25.0^oC[/tex] to water at [tex]90^oC[/tex] is, 7.712 KJ
Solution :
Process involved in the calculation of enthalpy change :
[tex](1):ice(-25^oC)\rightarrow ice(0^oC)\\\\(2):ice(0^oC)\rightarrow water(0^oC)\\\\(3):water(0^oC)\rightarrow water(90^oC)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=[m\times c_{ice}\times (T_2-T_1)]+\Delta H_{fusion}+[m\times c_{water}\times (T_3-T_2)][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change
m = mass of water = [tex]1mole\times 18g/mole=18g[/tex]
[tex]c_{ice}[/tex] = specific heat of ice = 2.09 J/gk
[tex]c_{water}[/tex] = specific heat of water = 4.18 J/gk
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole
conversion : [tex]0^oC=273k[/tex]
[tex]T_1[/tex] = initial temperature of ice = [tex]0^oC=273k[/tex]
[tex]T_2[/tex] = final temperature of ice = [tex]-25^oC=273+(-25)=248k[/tex]
[tex]T_3[/tex] = initial temperature of water = [tex]0^oC=273k[/tex]
[tex]T_4[/tex] = final temperature of water = [tex]90^oC=273+90=363k[/tex]
Now put all the given values in the above expression, we get
[tex]\Delta H=[18g\times 2.09J/gK\times (273-248)k]+0.00601J+[18g\times 4.18J/gK\times (363-273)k][/tex]
[tex]\Delta H=7712.106J=7.712KJ[/tex] (1 KJ = 1000 J)
Therefore, the enthalpy change for converting 1 mole of ice at [tex]-25.0^oC[/tex] to water at [tex]90^oC[/tex] is, 7.712 KJ
