Respuesta :
Answer : The heat absorbed by the ice cube and resulting water is, 9.15 kJ
Solution :
The process involved in this problem are :
[tex](1):H_2O(s)(-10.6^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(29.0^oC)[/tex]
The expression used will be:
[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = heat available for the reaction = [tex]4.50\times 10^3kJ=4.50\times 10^6J[/tex]
m = mass of ice = 19.2 g
[tex]c_{p,s}[/tex] = specific heat of solid water or ice = [tex]2.01J/g^oC[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = [tex]6.01kJ/mole=6010J/mole=\frac{6010J/mole}{18g/mole}J/g=333.89J/g[/tex]
Molar mass of water = 18 g/mole
Now put all the given values in the above expression, we get:
[tex]\Delta H=[19.2g\times 2.01J/g^oC\times (0-(-10.6))^oC]+19.2g\times 333.89J/g+[19.2g\times 4.18J/g^oC\times (29.0-0)^oC][/tex]
[tex]\Delta H=9147.1872J=9.15kJ[/tex]
Therefore, the heat absorbed by the ice cube and resulting water is, 9.15 kJ
