Certainly! Let's solve this problem step-by-step.
### 1. Understanding the Problem:
We have:
- Mass of the iron kettle, [tex]\( m_k \)[/tex] = 1.29 kg
- Mass of the water, [tex]\( m_w \)[/tex] = 2.93 kg
- Initial temperature, [tex]\( T_i \)[/tex] = 22.1°C
- Final temperature, [tex]\( T_f \)[/tex] = 71.2°C
- Specific heat capacity of iron, [tex]\( c_k \)[/tex] = 450 J/(kg·°C)
- Specific heat capacity of water, [tex]\( c_w \)[/tex] = 4186 J/(kg·°C)
### 2. Temperature Change:
First, we determine the change in temperature ([tex]\( \Delta T \)[/tex]) for both the kettle and the water.
[tex]\[ \Delta T = T_f - T_i = 71.2°C - 22.1°C = 49.1°C \][/tex]
### 3. Calculating Energy Absorbed by the Kettle:
The energy absorbed ([tex]\( Q \)[/tex]) is calculated using the formula:
[tex]\[ Q = mc\Delta T \][/tex]
For the kettle:
[tex]\[ Q_k = m_k \cdot c_k \cdot \Delta T \][/tex]
[tex]\[ Q_k = 1.29\, \text{kg} \times 450\, \text{J/(kg·°C)} \times 49.1\, \text{°C} \][/tex]
[tex]\[ Q_k = 1.29 \times 450 \times 49.1 \][/tex]
[tex]\[ Q_k = 28424.55\, \text{J} \][/tex]
### 4. Calculating Energy Absorbed by the Water:
For the water:
[tex]\[ Q_w = m_w \cdot c_w \cdot \Delta T \][/tex]
[tex]\[ Q_w = 2.93\, \text{kg} \times 4186\, \text{J/(kg·°C)} \times 49.1\, \text{°C} \][/tex]
[tex]\[ Q_w = 2.93 \times 4186 \times 49.1 \][/tex]
[tex]\[ Q_w = 601241.578\, \text{J} \][/tex]
### 5. Total Energy Absorbed:
The total energy absorbed by both the kettle and the water is the sum of the energy absorbed by each.
[tex]\[ Q_{total} = Q_k + Q_w \][/tex]
[tex]\[ Q_{total} = 28424.55\, \text{J} + 601241.578\, \text{J} \][/tex]
[tex]\[ Q_{total} = 629666.128\, \text{J} \][/tex]
### Final Answer:
- Energy absorbed by the kettle: [tex]\( 28424.55 \, \text{J} \)[/tex]
- Energy absorbed by the water: [tex]\( 601241.578 \, \text{J} \)[/tex]
- Total energy absorbed: [tex]\( 629666.128 \, \text{J} \)[/tex]
These values represent the amount of energy absorbed by the kettle and the water when they are heated from 22.1°C to 71.2°C.
