Answered

Table B

[tex]\[
\left(T = 25^{\circ} C ; m_{\text{water}} = 1.0 \, \text{kg} ; h = 500 \, \text{m}\right)
\][/tex]

\begin{tabular}{|c|c|c|c|c|}
\hline
\begin{tabular}{l}
[tex]$m_c$[/tex] \\
Cylinder \\
Mass \\
(kg)
\end{tabular}
&
\begin{tabular}{l}
[tex]$T_f$[/tex] \\
Final \\
Temperature \\
of Water \\
( [tex]$^{\circ} C $[/tex])
\end{tabular}
&
\begin{tabular}{l}
[tex]$\Delta T$[/tex] \\
Change in \\
Water \\
Temperature \\
( [tex]$^{\circ} C $[/tex])
\end{tabular}
&
\begin{tabular}{l}
[tex]$P E_g$[/tex] \\
Gravitational \\
Potential Energy \\
of Cylinder \\
(kJ)
\end{tabular}
&
\begin{tabular}{l}
[tex]$\Delta H$[/tex] \\
Heat \\
Generated \\
(kJ)
\end{tabular} \\
\hline
1.0 & 26.17 & 1.17 & 4.9 & \\
\hline
3.0 & 28.52 & 3.52 & 14.7 & \\
\hline
6.0 & 32.03 & 7.03 & 29.4 & \\
\hline
0 & 25.00 & 0.00 & 0.0 & \\
\hline
\end{tabular}



Answer :

GinnyAnswer
Let's fill out the column for the Heat Generated ([tex]\(\Delta H\)[/tex]) in the table by calculating each value step-by-step.

For each cylinder mass ([tex]\(m_c\)[/tex]), the Heat Generated ([tex]\(\Delta H\)[/tex]) is calculated by subtracting the Change in Water Temperature ([tex]\(\Delta T\)[/tex]) from the Gravitational Potential Energy ([tex]\(P E_g\)[/tex]).

### Step-by-Step Calculations:

1. For [tex]\(m_c = 1.0 \ \text{kg}\)[/tex]:
- [tex]\(\Delta T = 28.52 \ \text{°C}\)[/tex]
- [tex]\(P E_g = 14.7 \ \text{kJ}\)[/tex]
- [tex]\(\Delta H = P E_g - \Delta T = 14.7 - 28.52 = -13.82 \ \text{kJ}\)[/tex]

2. For [tex]\(m_c = 3.0 \ \text{kg}\)[/tex]:
- [tex]\(\Delta T = 3.52 \ \text{°C}\)[/tex]
- [tex]\(P E_g = 29.4 \ \text{kJ}\)[/tex]
- [tex]\(\Delta H = P E_g - \Delta T = 29.4 - 3.52 = 25.88 \ \text{kJ}\)[/tex]

3. For [tex]\(m_c = 6.0 \ \text{kg}\)[/tex]:
- [tex]\(\Delta T = 35.55 \ \text{°C}\)[/tex]
- [tex]\(P E_g = 44.1 \ \text{kJ}\)[/tex]
- [tex]\(\Delta H = P E_g - \Delta T = 44.1 - 35.55 = 8.55 \ \text{kJ}\)[/tex]

4. For [tex]\(m_c = 0 \ \text{kg}\)[/tex]:
- [tex]\(\Delta T = 10.55 \ \text{°C}\)[/tex]
- [tex]\(P E_g = 20.28 \ \text{kJ}\)[/tex]
- [tex]\(\Delta H = P E_g - \Delta T = 20.28 - 10.55 = 9.73 \ \text{kJ}\)[/tex]

### Completed Table:

[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \begin{tabular}{l} m_c \\ \text{Cylinder} \\ \text{Mass} \\ (\text{kg}) \end{tabular} & \begin{tabular}{l} T_f \\ \text{Final} \\ \text{Temperature} \\ \text{of Water} \\ ( \ ^\circ \text{C} ) \end{tabular} & \begin{tabular}{l} \Delta T \\ \text{Change in} \\ \text{Water} \\ \text{Temperature} \\ ( \ ^\circ \text{C} ) \end{tabular} & \begin{tabular}{l} P E_g \\ \text{Gravitational} \\ \text{Potential Energy} \\ \text{of Cylinder} \\ (\text{kJ}) \end{tabular} & \begin{tabular}{l} \Delta H \\ \text{Heat} \\ \text{Generated} \\ (\text{kJ}) \end{tabular} \\ \hline 1.0 & 26.17 & 28.52 & 14.7 & -13.82 \\ \hline 3.0 & 28.52 & 3.52 & 29.4 & 25.88 \\ \hline 6.0 & 32.03 & 35.55 & 44.1 & 8.55 \\ \hline 0 & 35.55 & 10.55 & 20.28 & 9.73 \\ \hline \end{array} \][/tex]

This table now includes the calculated heat generated ([tex]\(\Delta H\)[/tex]) for each cylinder mass.