Respuesta :
Answer : The heat energy absorbed will be, [tex]1.23\times 10^5cal[/tex]
Solution :
The process involved in this problem are :
[tex](1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(32.0^oC)[/tex]
The expression used will be:
[tex]\Delta H=m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]
where,
m = mass of ice = 1100 g
[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=1100g\times 333.89J/g+[1100g\times 4.18J/g^oC\times (32.0-0)^oC][/tex]
[tex]\Delta H=514415J=122948.1358cal=1.23\times 10^5cal[/tex]
Conversion used : (1 cal = 4.184 J)
Therefore, the heat energy absorbed will be, [tex]1.23\times 10^5cal[/tex]
