To solve this problem, we need to calculate the total heat lost by 1.75 moles of water as it cools from $140^\circ C$ (in the gaseous state) to $-10^\circ C$ (in the solid state). This involves several steps because the heat change depends on phase transitions and specific heat capacities for different phases of water.

Steps:

1. Cool water vapor from $140^\circ C$ to $100^\circ C$:
   Use the specific heat capacity for gaseous water, $C_{\text{H}_2\text{O (g)}} = 33.1 \, \text{J/mol·K}$.

2. Condense water vapor at $100^\circ C$:
   The enthalpy of vaporization is $\Delta H_{\text{vap}} = 40.7 \, \text{kJ/mol}$.

3. Cool liquid water from $100^\circ C$ to $0^\circ C$:
   Use the specific heat capacity for liquid water, $C_{\text{H}_2\text{O (l)}} = 75.4 \, \text{J/mol·K}$.

4. Freeze liquid water at $0^\circ C$:
   The enthalpy of fusion is $\Delta H_{\text{fus}} = 6.02 \, \text{kJ/mol}$.

5. Cool solid water from $0^\circ C$ to $-10^\circ C$:
   Use the specific heat capacity for solid water, $C_{\text{H}_2\text{O (s)}} = 37.6 \, \text{J/mol·K}$.

Calculation:

1. Cool water vapor from $140^\circ C$ to $100^\circ C$:

$q_1 = n \cdot C_{\text{H}_2\text{O (g)}} \cdot \Delta T = 1.75 \, \text{mol} \cdot 33.1 \, \text{J/mol·K} \cdot (140 - 100) \, \text{K}$ $q_1 = 1.75 \cdot 33.1 \cdot 40 = 2317 \, \text{J} = 2.317 \, \text{kJ}$

2. Condense water vapor at $100^\circ C$:

$q_2 = n \cdot \Delta H_{\text{vap}} = 1.75 \, \text{mol} \cdot 40.7 \, \text{kJ/mol}$ $q_2 = 71.225 \, \text{kJ}$

3. Cool liquid water from $100^\circ C$ to $0^\circ C$:

$q_3 = n \cdot C_{\text{H}_2\text{O (l)}} \cdot \Delta T = 1.75 \, \text{mol} \cdot 75.4 \, \text{J/mol·K} \cdot (100 - 0) \, \text{K}$ $q_3 = 1.75 \cdot 75.4 \cdot 100 = 13,195 \, \text{J} = 13.195 \, \text{kJ}$

4. Freeze liquid water at $0^\circ C$:

$q_4 = n \cdot \Delta H_{\text{fus}} = 1.75 \, \text{mol} \cdot 6.02 \, \text{kJ/mol}$ $q_4 = 10.535 \, \text{kJ}$

5. Cool solid water from $0^\circ C$ to $-10^\circ C$:

$q_5 = n \cdot C_{\text{H}_2\text{O (s)}} \cdot \Delta T = 1.75 \, \text{mol} \cdot 37.6 \, \text{J/mol·K} \cdot (0 - (-10)) \, \text{K}$ $q_5 = 1.75 \cdot 37.6 \cdot 10 = 658 \, \text{J} = 0.658 \, \text{kJ}$

Total Heat Lost:

$q_{\text{total}} = q_1 + q_2 + q_3 + q_4 + q_5$ $q_{\text{total}} = 2.317 + 71.225 + 13.195 + 10.535 + 0.658$ $q_{\text{total}} = 97.93 \, \text{kJ}$

Final Answer:

The total heat lost is 97.93 kJ.

Would you like further clarification or a breakdown of any step?

Related Questions:

1. What happens to the entropy of the system during the phase transitions?
2. How would the calculations change if the water was initially a liquid at $140^\circ C$?
3. What is the significance of latent heat during phase changes?
4. Why does water have different specific heat capacities in different states?
5. How would the result change if only 1 mole of water was used?

Tip:

Always ensure that the units are consistent throughout your calculations to avoid errors. For example, convert $\text{J}$ to $\text{kJ}$ when summing heat energies.