Answer :
Answer:- 57.7 kJ of heat is evolved.
Solution:- The conversion of 140 degree C steam to -55 degree C ice takes place in steps as:
First step:- Boiling point of water is 100 degree C, so first of all 140 degree C steam changes to 100 degree C steam. The heat for this step is calculated by using the formula, [tex]q=m*c*\Delta T[/tex]
where, q is the heat energy, m is mass in grams, c is heat capacity in [tex]\frac{J}{g.0_C}[/tex] and [tex]\Delta T[/tex] is change in temperature.
For first step, [tex]\Delta T[/tex] = 100 - 140 = -40 degree C
We have 1.00 mol that is 18.0 grams of steam and it's heat capacity is 2.01.
Let's calculate q for this step:
[tex]q_1=18.0*2.01(-40))[/tex]
[tex]q_1[/tex] = -1447.2 J
Second step:- In second step, 100 degree C steam changes to 100 degree C water. It's a phase change step. The equation used for this step is,
[tex]q=m*\\Delta H_V_a_p[/tex]
where, [tex]\Delta H_V_a_p[/tex] is the enthalpy of vaporization. It's value is 2260 J per gram.
Heat is released in this step as steam to water conversion is exothermic.
[tex]q_2=18.0(2260)[/tex]
[tex]q_2[/tex] = 40680 J
Since heat is released for steam to water conversion, the value of q for this second step is -40680 J.
Third step:- 100 degree C water is converted to 0 degree C water as 0 degree C is the melting point or freezing point for water. heat capacity for water is 4.18. Temperature change for this step is, = 0 - 100 = -100
[tex]q_3=18.0(4.18)(-100)[/tex]
[tex]q_3[/tex] = -7524 J
Fourth step:- 0 degree C water is converted to 0 degree C ice. Enthalpy of fusion for ice is 333.55 J per g. let's calculate the q for this step as:
[tex]q_4=m*\Delta H_f_u_s[/tex]
[tex]q_4[/tex] = 18.0(333.55)
[tex]q_4[/tex] = 6003.9 J
Since the heat is released, the energy for this step is also negative and it is -6003.9 J.
Fifth step:- 0 degree C ice is converted to -55.0 degree C.
[tex]\Delta T[/tex] = -55.0 - 0 = -55.0
[tex]q_5=18.0(2.09)(-55.0)[/tex]
[tex]q_5[/tex] = -2069.1 J
Total heat for the whole process is the sum of q values of all the five steps.
q=-1447.2J+(-40680J)+(-7524J)+(-6003.9J)+(-2069.1J)
q = -57724.2 J or -57.7 kJ
Negative sign indicates the heat is released.
So, 57.7 kJ of heat is released when 1.00 mol of 140.0 degree C of steam is converted to -55.0 degree C of ice.