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Let's walk through the step-by-step solution for determining the potential energy and the change in enthalpy for each height. The data provided includes heights ([tex]$h$[/tex]), final temperatures ([tex]$T_f$[/tex]), and changes in temperature ([tex]$\Delta T$[/tex]). We'll use the mass of the water ([tex]$m_w = 1.0 \text{ kg}$[/tex]), the mass of the cylinder ([tex]$m_c = 5.0 \text{ kg}$[/tex]), and the gravitational acceleration ([tex]$g = 9.81 \text{ m/s}^2$[/tex]) to calculate the potential energy at each height.
Step 1: Calculate the potential energy [tex]\( P E_g \)[/tex] for each height.
To find the potential energy, we use the formula:
[tex]\[ P E_g = (m_w + m_c) \cdot g \cdot h \][/tex]
1. For [tex]\( h = 100 \text{ m} \)[/tex]:
[tex]\[ P E_g = (1.0 \text{ kg} + 5.0 \text{ kg}) \cdot 9.81 \text{ m/s}^2 \cdot 100 \text{ m} = 6 \text{ kg} \cdot 9.81 \text{ m/s}^2 \cdot 100 \text{ m} = 5886 \text{ J} = 5.9 \text{ kJ} \][/tex]
2. For [tex]\( h = 200 \text{ m} \)[/tex]:
[tex]\[ P E_g = (1.0 \text{ kg} + 5.0 \text{ kg}) \cdot 9.81 \text{ m/s}^2 \cdot 200 \text{ m} = 6 \text{ kg} \cdot 9.81 \text{ m/s}^2 \cdot 200 \text{ m} = 11772 \text{ J} = 11.8 \text{ kJ} \][/tex]
3. For [tex]\( h = 500 \text{ m} \)[/tex]:
[tex]\[ P E_g = (1.0 \text{ kg} + 5.0 \text{ kg}) \cdot 9.81 \text{ m/s}^2 \cdot 500 \text{ m} = 6 \text{ kg} \cdot 9.81 \text{ m/s}^2 \cdot 500 \text{ m} = 29430 \text{ J} = 29.4 \text{ kJ} \][/tex]
4. For [tex]\( h = 1000 \text{ m} \)[/tex]:
[tex]\[ P E_g = (1.0 \text{ kg} + 5.0 \text{ kg}) \cdot 9.81 \text{ m/s}^2 \cdot 1000 \text{ m} = 6 \text{ kg} \cdot 9.81 \text{ m/s}^2 \cdot 1000 \text{ m} = 58860 \text{ J} = 58.9 \text{ kJ} \][/tex]
Step 2: Fill in the potential energy values [tex]\( P E_g \)[/tex] in the table.
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline h \, (m) & T_f \, (^\circ C) & \Delta T \, (^\circ C) & P E_g \, (kJ) & \Delta H \, (kJ) \\ \hline 100 & 26.17 & 1.17 & 5.9 & \\ \hline 200 & 27.34 & 2.34 & 11.8 & \\ \hline 500 & 30.86 & 5.86 & 29.4 & \\ \hline 1000 & 36.72 & 11.72 & 58.9 & \\ \hline \end{array} \][/tex]
Step 3: Calculate the change in enthalpy [tex]\( \Delta H \)[/tex].
The change in enthalpy [tex]\( \Delta H \)[/tex] can be considered equivalent to the potential energy [tex]\( P E_g \)[/tex] since they represent the energy available for heating the system:
[tex]\[ \Delta H = P E_g \][/tex]
Complete the table:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline h \, (m) & T_f \, (^\circ C) & \Delta T \, (^\circ C) & P E_g \, (kJ) & \Delta H \, (kJ) \\ \hline 100 & 26.17 & 1.17 & 5.9 & 5.9 \\ \hline 200 & 27.34 & 2.34 & 11.8 & 11.8 \\ \hline 500 & 30.86 & 5.86 & 29.4 & 29.4 \\ \hline 1000 & 36.72 & 11.72 & 58.9 & 58.9 \\ \hline \end{array} \][/tex]
Thus, the amount of heat generated for each height can simply be read off the [tex]\( \Delta H \)[/tex] column in the completed table, providing the rounded values for each height.
Step 1: Calculate the potential energy [tex]\( P E_g \)[/tex] for each height.
To find the potential energy, we use the formula:
[tex]\[ P E_g = (m_w + m_c) \cdot g \cdot h \][/tex]
1. For [tex]\( h = 100 \text{ m} \)[/tex]:
[tex]\[ P E_g = (1.0 \text{ kg} + 5.0 \text{ kg}) \cdot 9.81 \text{ m/s}^2 \cdot 100 \text{ m} = 6 \text{ kg} \cdot 9.81 \text{ m/s}^2 \cdot 100 \text{ m} = 5886 \text{ J} = 5.9 \text{ kJ} \][/tex]
2. For [tex]\( h = 200 \text{ m} \)[/tex]:
[tex]\[ P E_g = (1.0 \text{ kg} + 5.0 \text{ kg}) \cdot 9.81 \text{ m/s}^2 \cdot 200 \text{ m} = 6 \text{ kg} \cdot 9.81 \text{ m/s}^2 \cdot 200 \text{ m} = 11772 \text{ J} = 11.8 \text{ kJ} \][/tex]
3. For [tex]\( h = 500 \text{ m} \)[/tex]:
[tex]\[ P E_g = (1.0 \text{ kg} + 5.0 \text{ kg}) \cdot 9.81 \text{ m/s}^2 \cdot 500 \text{ m} = 6 \text{ kg} \cdot 9.81 \text{ m/s}^2 \cdot 500 \text{ m} = 29430 \text{ J} = 29.4 \text{ kJ} \][/tex]
4. For [tex]\( h = 1000 \text{ m} \)[/tex]:
[tex]\[ P E_g = (1.0 \text{ kg} + 5.0 \text{ kg}) \cdot 9.81 \text{ m/s}^2 \cdot 1000 \text{ m} = 6 \text{ kg} \cdot 9.81 \text{ m/s}^2 \cdot 1000 \text{ m} = 58860 \text{ J} = 58.9 \text{ kJ} \][/tex]
Step 2: Fill in the potential energy values [tex]\( P E_g \)[/tex] in the table.
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline h \, (m) & T_f \, (^\circ C) & \Delta T \, (^\circ C) & P E_g \, (kJ) & \Delta H \, (kJ) \\ \hline 100 & 26.17 & 1.17 & 5.9 & \\ \hline 200 & 27.34 & 2.34 & 11.8 & \\ \hline 500 & 30.86 & 5.86 & 29.4 & \\ \hline 1000 & 36.72 & 11.72 & 58.9 & \\ \hline \end{array} \][/tex]
Step 3: Calculate the change in enthalpy [tex]\( \Delta H \)[/tex].
The change in enthalpy [tex]\( \Delta H \)[/tex] can be considered equivalent to the potential energy [tex]\( P E_g \)[/tex] since they represent the energy available for heating the system:
[tex]\[ \Delta H = P E_g \][/tex]
Complete the table:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline h \, (m) & T_f \, (^\circ C) & \Delta T \, (^\circ C) & P E_g \, (kJ) & \Delta H \, (kJ) \\ \hline 100 & 26.17 & 1.17 & 5.9 & 5.9 \\ \hline 200 & 27.34 & 2.34 & 11.8 & 11.8 \\ \hline 500 & 30.86 & 5.86 & 29.4 & 29.4 \\ \hline 1000 & 36.72 & 11.72 & 58.9 & 58.9 \\ \hline \end{array} \][/tex]
Thus, the amount of heat generated for each height can simply be read off the [tex]\( \Delta H \)[/tex] column in the completed table, providing the rounded values for each height.
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