Respuesta :
Answer : The mass of ice is [tex]9.95\times 10^3g[/tex]
Solution :
The process involved in this problem are :
[tex](1):H_2O(s)(-19.9^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)(18.7^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 = ?
[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]4.50\times 10^6J=[m\times 2.01J/g^oC\times (0-(-19.9))^oC]+m\times 333.89J/g+[m\times 4.18J/g^oC\times (18.7-0)^oC][/tex]
[tex]m=9954.54093g=9.95\times 10^3g[/tex]
Therefore, the mass of ice is [tex]9.95\times 10^3g[/tex]
