Question

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 \(=2.02 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) $$\begin{array}{ll}\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta H_{\text {fusion }}=6.02 \mathrm{kJ} / \mathrm{mol}\left(\mathrm{at} 0^{\circ} \mathrm{C}\right) \\\\\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H_{\text {vaporization }}=40.7 \mathrm{kJ} / \mathrm{mol}\left(\text { at } 100 .^{\circ} \mathrm{C}\right)\end{array}$$

Short answer

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.

Step-by-step solution

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: \(q_{1} = 18.02 \mathrm{g} \times 2.03 \mathrm{J/^{\circ}C \cdot g} \times (0 - (-30))^{\circ}\mathrm{C} = 1095.81 \mathrm{J}\)

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: \(q_{2} = 6.02 \cdot 10^3\, \mathrm{J}\)

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: \(q_{3} = 18.02 \, \mathrm{g} \times 4.18 \, \mathrm{J/^{\circ}C\cdot g} \times (100 - 0)\,^{\circ}\mathrm{C} = 7534.44 \, \mathrm{J}\)

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: \(q_{4} = 40.7 \cdot 10^3\, \mathrm{J}\)

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.

Key concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept in thermochemistry that measures how much heat energy is needed to raise the temperature of a substance. It’s like understanding how much energy your car needs to go faster. For water, this value varies based on its state: solid (ice), liquid, or gas (steam).

• For ice: The specific heat capacity is given as 2.03 J/°C·g, meaning it takes 2.03 joules to raise the temperature of 1 gram of ice by 1°C.
• For liquid water: It is higher, at 4.18 J/°C·g, reflecting the fact that water can absorb more heat energy before its temperature rises compared to ice. This property is why water is good at moderating temperatures.
• For steam: The specific heat capacity is slightly less than that of ice, at 2.02 J/°C·g, because gaseous molecules already have higher energy levels and thus require slightly less additional energy per gram for the same temperature increase compared to liquid.

To use specific heat capacity in calculations, multiply it by the mass of the substance and the desired temperature change. This tells you the total heat energy required for that temperature change.

Heat of Fusion

The heat of fusion is the amount of energy required to change a substance from a solid to a liquid at its melting point without changing its temperature. Consider it the energy needed to "break up" the structure of ice to turn it into liquid water. For water, the heat of fusion is 6.02 kJ/mol.
This value means that for every mole of ice at 0°C that melts into water, 6.02 kJ of energy is required. Without this energy, the ice remains in its solid form despite the temperature being at the melting point.
In practical terms, if you have 1 mole of ice, you need to provide 6.02 kJ of energy just to change its state to a liquid, which we translate into 6,020 J for calculations. This energy input doesn't increase the water's temperature; rather, it's used entirely in changing the state from solid to liquid. Consequently, this is an important step when calculating the heat energy needed in a process where a change of state is involved.

Heat of Vaporization

The heat of vaporization is similar to the heat of fusion but involves changing a liquid to a gas at the boiling point. It represents the energy necessary to "evaporate" or vaporize a liquid into its gaseous form without changing its temperature. For water, the heat of vaporization is a significant 40.7 kJ/mol.
This means that for each mole of water at 100°C, turning it into steam requires 40.7 kJ of energy. In other words, this amount of energy is needed to overcome the intermolecular forces holding the liquid together and allow the molecules to freely disperse as a gas.
In terms of calculations, this energy amount translates to 40,700 J for 1 mole. When you compute the total energy required in a process that involves boiling, this heat of vaporization must be included. It's important when the transition involves moving past the boiling point, emphasizing why boiling requires so much energy, a fact that is evident when heating water on a stovetop.