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. $$ \begin{aligned} \text { Specific heat capacity of ice } &=2.03 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g} \\ \text { Specific heat capacity of water } &=4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g} \\ \text { Specific heat capacity of steam } &=2.02 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g} \end{aligned} $$ $$ \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{\mathrm{fision}}=6.02 \mathrm{kJ} / \mathrm{mol}\left(\mathrm{at} 0^{\circ} \mathrm{C}\right) $$ $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\mathrm{vaporization}}=40.7 \mathrm{kJ} / \mathrm{mol}\left(\mathrm{at} 100 .^{\circ} \mathrm{C}\right) $$

Short answer

The total heat, \(q\), required for the entire process can be calculated by following the steps: 1. Heat ice from -30°C to 0°C: \(q_1 = (1 \cdot 18.02\,g)(2.03\,\frac{J}{g\cdot °C})(0 - (-30))\,°C\) 2. Melt ice at 0°C: \(q_2 = (1\,mol)(6.02\,\frac{kJ}{mol})\) 3. Heat water from 0°C to 100°C: \(q_3 = (1 \cdot 18.02\,g)(4.18\,\frac{J}{g\cdot °C})(100 - 0)\,°C\) 4. Evaporate water at 100°C: \(q_4 = (1\,mol)(40.7\,\frac{kJ}{mol})\) 5. Heat steam from 100°C to 140°C: \(q_5 = (1 \cdot 18.02\,g)(2.02\,\frac{J}{g\cdot °C})(140 - 100)\,°C\) Add the heat calculated in each of the five steps to find the total heat required for the entire process: \(q = q_1 + q_2 + q_3 + q_4 + q_5\)

Step-by-step solution

Heating ice to 0°C

First, calculate the heat required to heat the ice from -30°C to 0°C. Use the specific heat capacity of ice and the mass of water, which is 1 mole × molar mass of water (18.02 g/mol). \(q_1 = mc\Delta T\) Where: \(m = 1 \cdot 18.02\,g\) \(c = 2.03\,\frac{J}{g\cdot °C}\) \(\Delta T = (0 - (-30))\,°C\) Calculate heat required, \(q_1\).

Melting ice

Next, calculate the heat needed to melt the ice at 0°C. Use the molar enthalpy of fusion. \(q_2 = n\Delta H_{fusion}\) Where: \(n = 1\,mol\) \(\Delta H_{fusion} = 6.02\,\frac{kJ}{mol}\) Convert \(\Delta H_{fusion}\) to J/mol for consistency. Calculate heat required, \(q_2\).

Heating water to 100°C

Calculate the heat needed to heat the liquid water from 0°C to 100°C. Use the specific heat capacity of water and mass of water. \(q_3 = mc\Delta T\) Where: \(m = 1 \cdot 18.02\,g\) \(c = 4.18\,\frac{J}{g\cdot °C}\) \(\Delta T = (100 - 0)\,°C\) Calculate heat required, \(q_3\).

Evaporating water

Next, calculate the heat required to evaporate the water at 100°C. Use the molar enthalpy of vaporization. \(q_4 = n\Delta H_{vaporization}\) Where: \(n = 1\,mol\) \(\Delta H_{vaporization} = 40.7\,\frac{kJ}{mol}\) Convert \(\Delta H_{vaporization}\) to J/mol for consistency. Calculate heat required, \(q_4\).

Heating steam to 140°C

Lastly, calculate the heat required to heat the steam from 100°C to 140°C. Use the specific heat capacity of steam and mass of water. \(q_5 = mc\Delta T\) Where: \(m = 1 \cdot 18.02\,g\) \(c = 2.02\,\frac{J}{g\cdot °C}\) \(\Delta T = (140 - 100)\,°C\) Calculate heat required, \(q_5\).

Total heat for the entire process

Calculate the total heat, \(q\), required for the entire process by adding the heat calculated in each of the five steps. \(q = q_1 + q_2 + q_3 + q_4 + q_5\)

Key concepts

Specific Heat Capacity

Specific heat capacity is a measure of the amount of heat required to change the temperature of a substance by one degree Celsius. It plays an essential role in calculating energy changes as a substance moves through different temperature ranges. When dealing with different states of matter, such as ice, water, and steam, each has its own specific heat capacity. For example, the specific heat capacity of ice is given as 2.03 J/g°C, which means for every gram of ice, you need 2.03 joules to raise its temperature by one degree Celsius. Similarly, water and steam have their specific heat capacities, 4.18 J/g°C and 2.02 J/g°C, respectively.

These values are crucial for calculating the amount of energy needed during various stages of heating. By applying the formula \(q = mc\Delta T\), where \(q\) is the heat added, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature, one can determine the heat required to warm the ice, liquid water, or steam by a specific amount.

Phase Transitions

Phase transitions occur when a substance changes from one state of matter to another, such as from solid to liquid, or liquid to gas. During a phase transition, the temperature remains constant, even though energy is being added or removed. This is because the energy goes into breaking or forming the bonds that hold the molecules in a particular phase, rather than increasing the temperature.

For example, when ice melts into water at 0°C, and water vaporizes into steam at 100°C, specific amounts of energy are required to facilitate these changes. This energy is related to molar enthalpy of fusion for melting ice and molar enthalpy of vaporization for boiling water. Understanding phase transitions is key to solving problems involving enthalpy calculations, as they often require calculating the energy needed for a substance to transform between different phases.

Molar Enthalpy of Fusion

Molar enthalpy of fusion is the amount of energy required to change one mole of a substance from a solid to a liquid at its melting point. In the context of water, the molar enthalpy of fusion is 6.02 kJ/mol, which reflects the energy needed to melt ice at 0°C into liquid water without changing its temperature.

To calculate this energy change, the formula \(q_2 = n\Delta H_{fusion}\) is used, where \(n\) is the number of moles and \(\Delta H_{fusion}\) is the molar enthalpy of fusion. This concept is critical in determining the energy changes that occur specifically during the melting process of a substance and should be considered when calculating the total energy required to transform a substance through different phases and temperature ranges.

Molar Enthalpy of Vaporization

Molar enthalpy of vaporization refers to the energy required to convert one mole of a liquid into vapor at its boiling point. For water, the molar enthalpy of vaporization is 40.7 kJ/mol, indicating the substantial energy needed to change water at 100°C into steam, while its temperature remains constant.

During this process, the formula \(q_4 = n\Delta H_{vaporization}\) is applied, where \(n\) is the number of moles and \(\Delta H_{vaporization}\) is the molar enthalpy of vaporization. This calculation is crucial for understanding the energy dynamics during the boiling process. It's a key aspect of enthalpy calculations, especially in problems like the one at hand, where you turn water into steam, necessitating an energy input to overcome intermolecular forces and facilitate the phase change.