Answer :
To determine the amount of energy required to raise the temperature of 3 kg of lead from 15°C to 20°C, we will use the formula for specific heat capacity:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the energy required (in joules, J),
- [tex]\( m \)[/tex] is the mass of the substance (in grams, g),
- [tex]\( c \)[/tex] is the specific heat capacity of the substance (in J/g·°C),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
Let's go through the solution step-by-step:
1. Convert the mass of lead to grams:
- The mass given is 3 kg.
- Since 1 kg = 1000 g:
[tex]\[ m = 3 \text{ kg} \times 1000 \frac{\text{g}}{\text{kg}} = 3000 \text{ g} \][/tex]
2. Identify the specific heat capacity of lead:
- From the table, the specific heat capacity ([tex]\( c \)[/tex]) of lead is 0.129 J/g·°C.
3. Determine the change in temperature ([tex]\( \Delta T \)[/tex]):
- The initial temperature ([tex]\( T_i \)[/tex]) is 15°C.
- The final temperature ([tex]\( T_f \)[/tex]) is 20°C.
- Therefore, [tex]\(\Delta T = T_f - T_i = 20^\circ\text{C} - 15^\circ\text{C} = 5^\circ\text{C}\)[/tex].
4. Calculate the energy required (Q):
- Substitute the values into the formula:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
[tex]\[ Q = 3000 \text{ g} \times 0.129 \frac{\text{J}}{\text{g} \cdot ^\circ\text{C}} \times 5^\circ\text{C} \][/tex]
[tex]\[ Q = 3000 \times 0.129 \times 5 = 1935 \text{ J} \][/tex]
Therefore, the energy required to raise the temperature of 3 kg of lead from 15°C to 20°C is 1935 joules (J).
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the energy required (in joules, J),
- [tex]\( m \)[/tex] is the mass of the substance (in grams, g),
- [tex]\( c \)[/tex] is the specific heat capacity of the substance (in J/g·°C),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
Let's go through the solution step-by-step:
1. Convert the mass of lead to grams:
- The mass given is 3 kg.
- Since 1 kg = 1000 g:
[tex]\[ m = 3 \text{ kg} \times 1000 \frac{\text{g}}{\text{kg}} = 3000 \text{ g} \][/tex]
2. Identify the specific heat capacity of lead:
- From the table, the specific heat capacity ([tex]\( c \)[/tex]) of lead is 0.129 J/g·°C.
3. Determine the change in temperature ([tex]\( \Delta T \)[/tex]):
- The initial temperature ([tex]\( T_i \)[/tex]) is 15°C.
- The final temperature ([tex]\( T_f \)[/tex]) is 20°C.
- Therefore, [tex]\(\Delta T = T_f - T_i = 20^\circ\text{C} - 15^\circ\text{C} = 5^\circ\text{C}\)[/tex].
4. Calculate the energy required (Q):
- Substitute the values into the formula:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
[tex]\[ Q = 3000 \text{ g} \times 0.129 \frac{\text{J}}{\text{g} \cdot ^\circ\text{C}} \times 5^\circ\text{C} \][/tex]
[tex]\[ Q = 3000 \times 0.129 \times 5 = 1935 \text{ J} \][/tex]
Therefore, the energy required to raise the temperature of 3 kg of lead from 15°C to 20°C is 1935 joules (J).
