To calculate the amount of energy involved when water cools down from #105^@C# to #-2^@C#, we will consider the phase transitions that the water go through:

#H_2O(g, 105^@C)->H_2O(g, 100^@C)" " "q_1#

#H_2O(g, 100^@C)->H_2O(l, 100^@C)" " "q_2#

#H_2O(l, 100^@C)->H_2O(l, 0^@C)" " " "q_3#

#H_2O(l, 0^@C)->H_2O(s, 0^@C)" " " "q_4#

#H_2O(s, 0^@C)->H_2O(s, -2^@C)" " "q_5#

The overall equation will be:

#H_2O(g, 105^@C)->H_2O(s, -2^@C)" " "q_t#

where, #q_t=q_1+q_2+q_3+q_4+q_5#

Now we will need to calculate each heat evolved for each equation:

To find, #q_1# we will consider the heat involved using:

#q=mxxsxxDeltaT#

where, #q# is the amount of heat, #m# is the mass of water, #s=2.0J/(""^@C*g)# is the specific heat capacity of steam and #DeltaT=T_f-T_i# is the change on temperature.

To find #q_3#, we will consider the heat involved when converting water liquid from #100^@C# to #0^@C# using the specific heat capacity of water liquid (#s=4.18J/(""^@C*g)# is the specific heat capacity of water liquid):

To find #q_5#, we will consider the heat involved when converting water liquid from #0^@C# to #-2^@C# using the specific heat capacity of ice (#s=2.03J/(""^@C*g)# is the specific heat capacity of water liquid):