# Question a3360

Mar 10, 2016

$\Delta H = - 54.5 \frac{k J}{m o l}$

#### Explanation:

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

${H}_{2} O \left(g , {105}^{\circ} C\right) \to {H}_{2} O \left(g , {100}^{\circ} C\right) \text{ " } {q}_{1}$

${H}_{2} O \left(g , {100}^{\circ} C\right) \to {H}_{2} O \left(l , {100}^{\circ} C\right) \text{ " } {q}_{2}$

${H}_{2} O \left(l , {100}^{\circ} C\right) \to {H}_{2} O \left(l , {0}^{\circ} C\right) \text{ " " } {q}_{3}$

${H}_{2} O \left(l , {0}^{\circ} C\right) \to {H}_{2} O \left(s , {0}^{\circ} C\right) \text{ " " } {q}_{4}$

${H}_{2} O \left(s , {0}^{\circ} C\right) \to {H}_{2} O \left(s , - {2}^{\circ} C\right) \text{ " } {q}_{5}$

The overall equation will be:

${H}_{2} O \left(g , {105}^{\circ} C\right) \to {H}_{2} O \left(s , - {2}^{\circ} C\right) \text{ " } {q}_{t}$

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:

$q = m \times s \times \Delta T$

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 $\Delta T = {T}_{f} - {T}_{i}$ is the change on temperature.

$\implies \Delta T = 100 - 105 = - {5}^{\circ} C$

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 $\Delta {H}_{c o n d} = - 40.7 \frac{k J}{m o l}$ of water:

The number of mole of water is: $n = \frac{m}{M M} = \frac{4.0 \cancel{g}}{18 \frac{\cancel{g}}{m o l}} = 0.22 m o l$

Thus, ${q}_{2} = \frac{- 40.7 k J}{1 \cancel{m o l}} \times 0.22 \cancel{m o l} = - 9.0 k J = - 9000.0 J$

To find ${q}_{3}$, we will consider the heat involved when converting water liquid from ${100}^{\circ} C$ to ${0}^{\circ} C$ using the specific heat capacity of water liquid (s=4.18J/(""^@C*g) is the specific heat capacity of water liquid):

$q = m \times s \times \Delta T$

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 $\Delta {H}_{\text{freezing}} = - 6.02 \frac{k J}{m o l}$ of water:

Thus, ${q}_{4} = \frac{- 6.02 k J}{1 \cancel{m o l}} \times 0.22 \cancel{m o l} = - 1.32 k J = - 1320 J$

To find ${q}_{5}$, we will consider the heat involved when converting water liquid from ${0}^{\circ} C$ to $- {2}^{\circ} C$ using the specific heat capacity of ice (s=2.03J/(""^@C*g) is the specific heat capacity of water liquid):

$q = m \times s \times \Delta T$

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

Therefore, ${q}_{t} = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5} = - 40.0 J + \left(- 9000.0 J\right) + \left(- 1672.0 J\right) + \left(- 1320 J\right) + \left(- 16.0 J\right) = - 12048 = - 12.0 k J$

$\implies \Delta H = \frac{q}{n} = \frac{- 12.0}{0.22 m o l} = - 54.5 \frac{k J}{m o l}$

Thermochemistry | Enthalpy and Calorimetry.