Answer :
To calculate the amount of heat generated for each cylinder height, we need to consider the mass of water ([tex]\(m_w\)[/tex]), the mass of the cylinder ([tex]\(m_c\)[/tex]), and the specific heat capacities of water and the cylinder material. Additionally, we need to use the temperature changes ([tex]\(\Delta T\)[/tex]) provided in the table.
The formula for calculating the heat generated ([tex]\(\Delta H\)[/tex]) when there is a temperature change is:
[tex]\[ \Delta H = m \cdot c \cdot \Delta T \][/tex]
where [tex]\(m\)[/tex] is the mass, [tex]\(c\)[/tex] is the specific heat capacity, and [tex]\(\Delta T\)[/tex] is the change in temperature.
Given:
- Mass of water: [tex]\(m_w = 1.0 \, \text{kg}\)[/tex]
- Mass of the cylinder: [tex]\(m_c = 5.0 \, \text{kg}\)[/tex]
- Specific heat capacity of water: [tex]\(c_w = 4.18 \, \text{kJ/(kg} \cdot \text{°C)}\)[/tex]
- Specific heat capacity of the cylinder: [tex]\(c_c = 0.9 \, \text{kJ/(kg} \cdot \text{°C)}\)[/tex]
The temperature changes ([tex]\(\Delta T\)[/tex]) for different heights are provided in the table:
- [tex]\(\Delta T\)[/tex] for 100 m: [tex]\(1.17 \, \text{°C}\)[/tex]
- [tex]\(\Delta T\)[/tex] for 200 m: [tex]\(2.34 \, \text{°C}\)[/tex]
- [tex]\(\Delta T\)[/tex] for 500 m: [tex]\(5.86 \, \text{°C}\)[/tex]
- [tex]\(\Delta T\)[/tex] for 1000 m: [tex]\(11.72 \, \text{°C}\)[/tex]
Let's calculate the heat generated for each height:
1. For 100 meters:
[tex]\[ \Delta H_{100} = m_w \cdot c_w \cdot \Delta T_{100} + m_c \cdot c_c \cdot \Delta T_{100} \][/tex]
[tex]\[ \Delta H_{100} = 1.0 \, \text{kg} \cdot 4.18 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 1.17 \, \text{°C} + 5.0 \, \text{kg} \cdot 0.9 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 1.17 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{100} = 4.8906 \, \text{kJ} + 5.265 \, \text{kJ} = 10.2 \, \text{kJ} \ (\text{rounded}) \][/tex]
2. For 200 meters:
[tex]\[ \Delta H_{200} = m_w \cdot c_w \cdot \Delta T_{200} + m_c \cdot c_c \cdot \Delta T_{200} \][/tex]
[tex]\[ \Delta H_{200} = 1.0 \, \text{kg} \cdot 4.18 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 2.34 \, \text{°C} + 5.0 \, \text{kg} \cdot 0.9 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 2.34 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{200} = 9.7812 \, \text{kJ} + 10.53 \, \text{kJ} = 20.3 \, \text{kJ} \ (\text{rounded}) \][/tex]
3. For 500 meters:
[tex]\[ \Delta H_{500} = m_w \cdot c_w \cdot \Delta T_{500} + m_c \cdot c_c \cdot \Delta T_{500} \][/tex]
[tex]\[ \Delta H_{500} = 1.0 \, \text{kg} \cdot 4.18 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 5.86 \, \text{°C} + 5.0 \, \text{kg} \cdot 0.9 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 5.86 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{500} = 24.5148 \, \text{kJ} + 26.37 \, \text{kJ} = 50.9 \, \text{kJ} \ (\text{rounded}) \][/tex]
4. For 1000 meters:
[tex]\[ \Delta H_{1000} = m_w \cdot c_w \cdot \Delta T_{1000} + m_c \cdot c_c \cdot \Delta T_{1000} \][/tex]
[tex]\[ \Delta H_{1000} = 1.0 \, \text{kg} \cdot 4.18 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 11.72 \, \text{°C} + 5.0 \, \text{kg} \cdot 0.9 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 11.72 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{1000} = 48.9896 \, \text{kJ} + 52.74 \, \text{kJ} = 101.7 \, \text{kJ} \ (\text{rounded}) \][/tex]
Final answers:
- For 100 meters: [tex]\(10.2 \, \text{kJ}\)[/tex]
- For 200 meters: [tex]\(20.3 \, \text{kJ}\)[/tex]
- For 500 meters: [tex]\(50.9 \, \text{kJ}\)[/tex]
- For 1000 meters: [tex]\(101.7 \, \text{kJ}\)[/tex]
Therefore, the amount of heat generated for each height is as follows:
[tex]\[ \begin{array}{l} 100 \, \text{m}: 10.2 \, \text{kJ} \\ 200 \, \text{m}: 20.3 \, \text{kJ} \\ 500 \, \text{m}: 50.9 \, \text{kJ} \\ 1000 \, \text{m}: 101.7 \, \text{kJ} \end{array} \][/tex]
The formula for calculating the heat generated ([tex]\(\Delta H\)[/tex]) when there is a temperature change is:
[tex]\[ \Delta H = m \cdot c \cdot \Delta T \][/tex]
where [tex]\(m\)[/tex] is the mass, [tex]\(c\)[/tex] is the specific heat capacity, and [tex]\(\Delta T\)[/tex] is the change in temperature.
Given:
- Mass of water: [tex]\(m_w = 1.0 \, \text{kg}\)[/tex]
- Mass of the cylinder: [tex]\(m_c = 5.0 \, \text{kg}\)[/tex]
- Specific heat capacity of water: [tex]\(c_w = 4.18 \, \text{kJ/(kg} \cdot \text{°C)}\)[/tex]
- Specific heat capacity of the cylinder: [tex]\(c_c = 0.9 \, \text{kJ/(kg} \cdot \text{°C)}\)[/tex]
The temperature changes ([tex]\(\Delta T\)[/tex]) for different heights are provided in the table:
- [tex]\(\Delta T\)[/tex] for 100 m: [tex]\(1.17 \, \text{°C}\)[/tex]
- [tex]\(\Delta T\)[/tex] for 200 m: [tex]\(2.34 \, \text{°C}\)[/tex]
- [tex]\(\Delta T\)[/tex] for 500 m: [tex]\(5.86 \, \text{°C}\)[/tex]
- [tex]\(\Delta T\)[/tex] for 1000 m: [tex]\(11.72 \, \text{°C}\)[/tex]
Let's calculate the heat generated for each height:
1. For 100 meters:
[tex]\[ \Delta H_{100} = m_w \cdot c_w \cdot \Delta T_{100} + m_c \cdot c_c \cdot \Delta T_{100} \][/tex]
[tex]\[ \Delta H_{100} = 1.0 \, \text{kg} \cdot 4.18 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 1.17 \, \text{°C} + 5.0 \, \text{kg} \cdot 0.9 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 1.17 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{100} = 4.8906 \, \text{kJ} + 5.265 \, \text{kJ} = 10.2 \, \text{kJ} \ (\text{rounded}) \][/tex]
2. For 200 meters:
[tex]\[ \Delta H_{200} = m_w \cdot c_w \cdot \Delta T_{200} + m_c \cdot c_c \cdot \Delta T_{200} \][/tex]
[tex]\[ \Delta H_{200} = 1.0 \, \text{kg} \cdot 4.18 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 2.34 \, \text{°C} + 5.0 \, \text{kg} \cdot 0.9 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 2.34 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{200} = 9.7812 \, \text{kJ} + 10.53 \, \text{kJ} = 20.3 \, \text{kJ} \ (\text{rounded}) \][/tex]
3. For 500 meters:
[tex]\[ \Delta H_{500} = m_w \cdot c_w \cdot \Delta T_{500} + m_c \cdot c_c \cdot \Delta T_{500} \][/tex]
[tex]\[ \Delta H_{500} = 1.0 \, \text{kg} \cdot 4.18 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 5.86 \, \text{°C} + 5.0 \, \text{kg} \cdot 0.9 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 5.86 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{500} = 24.5148 \, \text{kJ} + 26.37 \, \text{kJ} = 50.9 \, \text{kJ} \ (\text{rounded}) \][/tex]
4. For 1000 meters:
[tex]\[ \Delta H_{1000} = m_w \cdot c_w \cdot \Delta T_{1000} + m_c \cdot c_c \cdot \Delta T_{1000} \][/tex]
[tex]\[ \Delta H_{1000} = 1.0 \, \text{kg} \cdot 4.18 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 11.72 \, \text{°C} + 5.0 \, \text{kg} \cdot 0.9 \, \text{kJ/(kg} \cdot \text{°C)} \cdot 11.72 \, \text{°C} \][/tex]
[tex]\[ \Delta H_{1000} = 48.9896 \, \text{kJ} + 52.74 \, \text{kJ} = 101.7 \, \text{kJ} \ (\text{rounded}) \][/tex]
Final answers:
- For 100 meters: [tex]\(10.2 \, \text{kJ}\)[/tex]
- For 200 meters: [tex]\(20.3 \, \text{kJ}\)[/tex]
- For 500 meters: [tex]\(50.9 \, \text{kJ}\)[/tex]
- For 1000 meters: [tex]\(101.7 \, \text{kJ}\)[/tex]
Therefore, the amount of heat generated for each height is as follows:
[tex]\[ \begin{array}{l} 100 \, \text{m}: 10.2 \, \text{kJ} \\ 200 \, \text{m}: 20.3 \, \text{kJ} \\ 500 \, \text{m}: 50.9 \, \text{kJ} \\ 1000 \, \text{m}: 101.7 \, \text{kJ} \end{array} \][/tex]
