IDNLearn.com provides a user-friendly platform for finding and sharing knowledge. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
Sagot :
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
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
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
