To determine the amount of heat generated ([tex]\(\Delta H\)[/tex]) for each cylinder height using the provided data, follow these steps:
1. Given Data:
- Initial temperature, [tex]\(T_i = 25^\circ \text{C}\)[/tex]
- Mass of water, [tex]\(m_w = 1.0 \text{kg}\)[/tex]
- Mass of cylinder, [tex]\(m_c = 5.0 \text{kg}\)[/tex]
- Specific heat capacity of water, [tex]\(c_w = 4.18 \text{ J/(g°C)} = 4.18 \times 10^{-3} \text{ kJ/(kg°C)}\)[/tex]
2. Data from the Table:
- Heights ([tex]\( h \)[/tex]): 100 m, 200 m, 500 m, 1000 m
- Final temperatures ([tex]\( T_f \)[/tex]): 26.17°C, 27.34°C, 30.86°C, 36.72°C
- Change in temperature ([tex]\( \Delta T \)[/tex]): 1.17°C, 2.34°C, 5.86°C, 11.72°C
3. Formula for Change in Enthalpy ([tex]\( \Delta H \)[/tex]):
[tex]\[ \Delta H = m_w \times c_w \times \Delta T \][/tex]
Here, [tex]\( m_w \)[/tex] is the mass of water (in kg), [tex]\( c_w \)[/tex] is the specific heat capacity of water (in kJ/(kg°C)), and [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
4. Calculations:
- For 100 m:
[tex]\[ \Delta H_{100} = 1.0 \, \text{kg} \times 4.18 \times 10^{-3} \, \text{kJ/(kg°C)} \times 1.17 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{100} = 0.0048906 \, \text{kJ} \][/tex]
- For 200 m:
[tex]\[ \Delta H_{200} = 1.0 \, \text{kg} \times 4.18 \times 10^{-3} \, \text{kJ/(kg°C)} \times 2.34 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{200} = 0.0097812 \, \text{kJ} \][/tex]
- For 500 m:
[tex]\[ \Delta H_{500} = 1.0 \, \text{kg} \times 4.18 \times 10^{-3} \, \text{kJ/(kg°C)} \times 5.86 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{500} = 0.0244948 \, \text{kJ} \][/tex]
- For 1000 m:
[tex]\[ \Delta H_{1000} = 1.0 \, \text{kg} \times 4.18 \times 10^{-3} \, \text{kJ/(kg°C)} \times 11.72 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{1000} = 0.0489896 \, \text{kJ} \][/tex]
5. Rounding to the Nearest Tenth:
- [tex]\(\Delta H_{100} \approx 0.0 \, \text{kJ}\)[/tex]
- [tex]\(\Delta H_{200} \approx 0.0 \, \text{kJ}\)[/tex]
- [tex]\(\Delta H_{500} \approx 0.0 \, \text{kJ}\)[/tex]
- [tex]\(\Delta H_{1000} \approx 0.0 \, \text{kJ}\)[/tex]
Thus, the calculations show very small values of generated heat when rounded to the nearest tenth place:
- For 100 m: approximately [tex]\(0.0\)[/tex] kJ
- For 200 m: approximately [tex]\(0.0\)[/tex] kJ
- For 500 m: approximately [tex]\(0.0\)[/tex] kJ
- For 1000 m: approximately [tex]\(0.0\)[/tex] kJ
