How much heat is evolved in converting 1.00 mol of steam at 140.0 ∘c to ice at -55.0 ∘c? the heat capacity of steam is 2.01 j/(g

Question

Ket [755]

3 years ago

10

How much heat is evolved in converting 1.00 mol of steam at 140.0 ∘c to ice at -55.0 ∘c? the heat capacity of steam is 2.01 j/(g

⋅∘c) and of ice is 2.09 j/(g⋅∘c)?

Chemistry

1 answer:

Sauron [17] · Answer 1 · 2022-03-12T13:31:51+03:00

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, $q=m*c*\Delta T$

where, q is the heat energy, m is mass in grams, c is heat capacity in $\frac{J}{g.0_C}$ and $\Delta T$ is change in temperature.

For first step, $\Delta T$ = 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:

$q_1=18.0*2.01(-40))$

$q_1$ = -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,

$q=m*\\Delta H_V_a_p$

where, $\Delta H_V_a_p$ 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.

$q_2=18.0(2260)$

$q_2$ = 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

$q_3=18.0(4.18)(-100)$

$q_3$ = -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:

$q_4=m*\Delta H_f_u_s$