You have a 1.00 -mole sample of water at \(-30 .^{\circ} \mathrm{C}\) and you heat it until you have gaseous water at \(140 .^{\circ} \mathrm{C}\). Calculate \(q\) for the entire process. Use the following data. Specific heat capacity of ice \(=2.03 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of water \(=4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of steam \(=2.02 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) $$\begin{array}{ll}\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta H_{\text {fusion }}=6.02 \mathrm{kJ} / \mathrm{mol}\left(\mathrm{at} 0^{\circ} \mathrm{C}\right) \\\\\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H_{\text {vaporization }}=40.7 \mathrm{kJ} / \mathrm{mol}\left(\text { at } 100 .^{\circ} \mathrm{C}\right)\end{array}$$

Step by step solution

01

1. Heat the ice to 0°C.

To heat the ice from -30°C to 0°C, use the formula \(q=m\cdot c\cdot \Delta T\), where m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature. First, we need to convert 1.00 mole of water into grams using its molar mass, which is \(18.02 \mathrm{g/mol}\). Mass of water: \(1.00 \mathrm{mol} \times 18.02 \mathrm{g/mol} = 18.02 \mathrm{g}\) Now, use the specific heat capacity of ice, which is \(2.03 \mathrm{J/^{\circ}C \cdot g}\). Heat energy to warm the ice: \(q_{1} = 18.02 \mathrm{g} \times 2.03 \mathrm{J/^{\circ}C \cdot g} \times (0 - (-30))^{\circ}\mathrm{C} = 1095.81 \mathrm{J}\)

02

2. Melt the ice at 0°C.

To melt the ice, we need to use the heat of fusion, which is given as ΔH_fusion = \(6.02 \mathrm{kJ/mol}\). Multiply the moles of water by the heat of fusion to obtain the required heat: Heat energy to melt the ice: \(q_{2} = 1.00 \, \mathrm{mol} \times 6.02 \, \mathrm{ kJ/mol} = 6.02 \, \mathrm{kJ}\) Convert to Joules: \(q_{2} = 6.02 \cdot 10^3\, \mathrm{J}\)

03

3. Heat the liquid water to 100°C.

Now, heat the liquid water from 0°C to 100°C, using the specific heat capacity of water, which is \(4.18 \mathrm{J/^{\circ}C\cdot g}\). Heat energy to warm the liquid water: \(q_{3} = 18.02 \, \mathrm{g} \times 4.18 \, \mathrm{J/^{\circ}C\cdot g} \times (100 - 0)\,^{\circ}\mathrm{C} = 7534.44 \, \mathrm{J}\)

04

4. Vaporize the liquid water at 100°C.

To vaporize the liquid water, use the heat of vaporization, which is given as ΔH_vaporization = \(40.7 \mathrm{kJ/mol}\). Multiply the moles of water by the heat of vaporization to obtain the required heat: Heat energy to vaporize the water: \(q_{4} = 1.00 \, \mathrm{mol} \times 40.7 \, \mathrm{ kJ/mol} = 40.7 \, \mathrm{kJ}\) Convert to Joules: \(q_{4} = 40.7 \cdot 10^3\, \mathrm{J}\)

05

5. Heat the steam to 140°C.

Finally, heat the steam from 100°C to 140°C, using the specific heat capacity of steam, which is \(2.02 \mathrm{J/^{\circ} C\cdot g}\). Heat energy to warm the steam: \(q_{5} = 18.02 \, \mathrm{g} \times 2.02 \, \mathrm{J/^{\circ}C\cdot g} \times (140 - 100)\,^{\circ}\mathrm{C} = 1623.61 \, \mathrm{J}\) Now, calculate the total heat energy required for the entire process by adding the heat energy calculated in each step: Total heat energy, q: \(q = q_{1} + q_{2} + q_{3} + q_{4} + q_{5} = 1095.81 \, \mathrm{J} + 6.02 \cdot 10^3\, \mathrm{J} + 7534.44 \, \mathrm{J} + 40.7 \cdot 10^3\, \mathrm{J} + 1623.61 \, \mathrm{J} = 58253.86\, \mathrm{J}\) So, the total heat energy required to heat a 1.00-mole sample of water from -30°C to gaseous water at 140°C is 58,253.86 J.

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