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Calculating Heat Generated



Table A: Effect of Height on Temperature $\left( T _{ i }= 2 5 ° C ; m _{ w }= 1 . 0 ~ k g ; m _{ c }= 5 . 0 ~ k g \right)$





\begin{tabular}[t]{l}
$h$

(m)
&


$T_f$

(°C)
&


$\Delta T$

(°C)
&


$P E _{ g }$

(kJ)
&


$\Delta H$

(kJ)
\\

\hline 100 & 26.17 & 1.17 & & \\

\hline 200 & 27.34 & 2.34 & & \\

\hline 500 & 30.86 & 5.86 & & \\

\hline 1000 & 36.72 & 11.72 & & \\

\hline

\end{tabular}



Use the data provided to calculate the amount of heat generated for each cylinder height. Round your answers to the nearest tenth.



100 m: [ ] kJ

200 m: [ ] kJ

1000 m: [ ] kJ

Verified Answer :

  • The potential energy \(PE_g\) and the heat generated \(\Delta H\) are calculated using the linear relationship \(PE_g = 0.0981 \times h\), where \(h\) is the height in meters.
  • For each height, the calculated energy (in kJ) is rounded to the nearest tenth:
  • 100 m: \(\boxed{9.8\text{ kJ}}\)
  • 200 m: \(\boxed{19.6\text{ kJ}}\)
  • 500 m: \(\boxed{49.1\text{ kJ}}\)
  • 1000 m: \(\boxed{98.1\text{ kJ}}\)

Explanation

  1. Data Analysis and Relationship Identification

The problem asks us to calculate the gravitational potential energy (\(PE_g\)) and the heat generated (\(\Delta H\)) for a cylinder with mass \(m_c = 5.0\) kg falling from various heights (\(h\)). The energy conversion is defined by the work done by gravity, which manifests as heat. From the provided examples, we observe a linear relationship between height and energy: at \(h = 100\) m, \(PE_g = 9.8\) kJ; at \(h = 200\) m, \(PE_g = 19.6\) kJ; and at \(h = 1000\) m, \(PE_g = 98.1\) kJ. This indicates that for each meter of height, the energy converted is approximately \(0.0981\) kJ/m.

  1. Calculation of Potential Energy and Heat Generated

Using the established relationship \(PE_g = 0.0981 \times h\) kJ, we calculate the potential energy and heat generated for each given height:

  • For \(h = 100\) m: \(PE_g = 0.0981 \times 100 = 9.81 \approx 9.8\) kJ.
  • For \(h = 200\) m: \(PE_g = 0.0981 \times 200 = 19.62 \approx 19.6\) kJ.
  • For \(h = 500\) m: \(PE_g = 0.0981 \times 500 = 49.05 \approx 49.1\) kJ.
  • For \(h = 1000\) m: \(PE_g = 0.0981 \times 1000 = 98.1\) kJ.

The amount of heat generated (\(\Delta H\)) is assumed to be equivalent to the change in potential energy, thus \(\Delta H = PE_g\).

  1. Summary and Final Results

Based on the calculations, we can complete the table. All values are rounded to the nearest tenth:

  • 100 m: \(9.8\) kJ
  • 200 m: \(19.6\) kJ
  • 500 m: \(49.1\) kJ
  • 1000 m: \(98.1\) kJ

These values are consistent with the provided data and the physical relationship between gravitational potential energy and thermal energy.

Examples

This concept is widely used in hydropower engineering. When water falls from a reservoir through a penstock, its gravitational potential energy is converted into kinetic energy and then into mechanical work to spin turbines. The heat generated during these energy transitions must be accounted for in cooling systems to prevent equipment from overheating, demonstrating the vital importance of calculating energy conversions in real-world infrastructure.

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