Question #a3360

1 Answer
Mar 10, 2016

Answer:

#DeltaH=-54.5(kJ)/(mol)#

Explanation:

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.

#=>DeltaT=100-105=-5^@C#

Thus, #q_1=4.0cancel(g)xx2.0J/(cancel(""^@C)*cancel(g))xx(-5)cancel(""^@C)=-40.0J#

To find #q_2#, we will use the #DeltaH_(cond)=-40.7(kJ)/(mol)# of water:

The number of mole of water is: #n=m/(MM)=(4.0cancel(g))/(18cancel(g)/(mol))=0.22mol#

Thus, #q_2=(-40.7kJ)/(1cancel(mol))xx0.22cancel(mol)=-9.0kJ=-9000.0J#

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):

#q=mxxsxxDeltaT#

Thus, #q_3=4.0cancel(g)xx4.18J/(cancel(""^@C)*cancel(g))xx(0-100)cancel(""^@C)=-1672.0J#

To find #q_4#, we will use the #DeltaH_("freezing")=-6.02(kJ)/(mol)# of water:

Thus, #q_4=(-6.02kJ)/(1cancel(mol))xx0.22cancel(mol)=-1.32kJ=-1320J#

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):

#q=mxxsxxDeltaT#

Thus, #q_5=4.0cancel(g)xx2.03J/(cancel(""^@C)*cancel(g))xx(-2-0)cancel(""^@C)=-16.0J#

Therefore, #q_t=q_1+q_2+q_3+q_4+q_5=-40.0J+(-9000.0J)+(-1672.0J)+(-1320J)+(-16.0J)=-12048=-12.0kJ#

#=>DeltaH=q/n=(-12.0)/(0.22mol)=-54.5(kJ)/(mol)#

Thermochemistry | Enthalpy and Calorimetry.