Answer :
To solve this problem, let's go through the calculations step by step. We will use the principle of conservation of energy, which states that the heat lost by water will be equal to the heat gained by the copper. In this case, we assume no heat is lost to the surroundings.
### Given Data:
1. Mass of Copper: 2 pounds
- Convert pounds to grams: [tex]\(2 \text{ pounds} \times 454 \text{ grams/pound} = 908 \text{ grams}\)[/tex]
2. Specific Heat of Copper: [tex]\(0.385 \text{ J/(g°C)}\)[/tex]
3. Initial Temperature of Copper (T_initial,Cu): 8°C
4. Final Temperature of System (T_final): 52°C
5. Initial Temperature of Water (T_initial,H2O): 60°C
6. Specific Heat of Water: [tex]\(4.186 \text{ J/(g°C)}\)[/tex]
### Step-by-Step Solution:
1. Calculate the Heat Energy Gained by Copper:
The heat energy (Q) gained or lost is given by the formula:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where
- [tex]\( m \)[/tex] is the mass
- [tex]\( c \)[/tex] is the specific heat capacity
- [tex]\( \Delta T \)[/tex] is the change in temperature
For copper:
- Mass ([tex]\( m_{Cu} \)[/tex]): 908 grams
- Change in temperature ([tex]\( \Delta T_{Cu} \)[/tex]): [tex]\( 52°C - 8°C = 44°C \)[/tex]
[tex]\[ Q_{Cu} = m_{Cu} \cdot c_{Cu} \cdot \Delta T_{Cu} \][/tex]
[tex]\[ Q_{Cu} = 908 \text{ g} \times 0.385 \text{ J/(g°C)} \times 44°C \][/tex]
[tex]\[ Q_{Cu} = 15409.36 \text{ J} \][/tex]
2. Set Up the Heat Lost by Water:
For water to cool from 60°C to 52°C, the heat lost can be calculated using the same formula. Assume the mass of water is [tex]\( m_{H2O} \)[/tex] grams (to find):
- Change in temperature ([tex]\( \Delta T_{H2O} \)[/tex]): [tex]\( 60°C - 52°C = 8°C \)[/tex]
[tex]\[ Q_{H2O} = m_{H2O} \cdot c_{H2O} \cdot \Delta T_{H2O} \][/tex]
[tex]\[ Q_{H2O} = m_{H2O} \cdot 4.186 \text{ J/(g°C)} \cdot 8°C \][/tex]
3. Equate the Heat Lost by Water to the Heat Gained by Copper:
[tex]\[ Q_{H2O} = Q_{Cu} \][/tex]
[tex]\[ m_{H2O} \cdot 4.186 \text{ J/(g°C)} \cdot 8°C = 15409.36 \text{ J} \][/tex]
4. Solve for the Mass of Water:
[tex]\[ m_{H2O} \cdot 4.186 \text{ J/(g°C)} \cdot 8°C = 15409.36 \text{ J} \][/tex]
[tex]\[ m_{H2O} \cdot 33.488 = 15409.36 \][/tex]
[tex]\[ m_{H2O} = \frac{15409.36}{33.488} \][/tex]
[tex]\[ m_{H2O} \approx 460.26 \text{ grams} \][/tex]
Since 1 milliliter (mL) of water has a mass of 1 gram, the volume of water needed is the same as its mass:
[tex]\[ \text{Volume of water} = 460.26 \text{ mL} \][/tex]
### Answer:
You would need approximately 460.26 milliliters of water at 60°C to get a final temperature of 52°C of the system after dropping in a 2-pound piece of copper that was at 8°C.
### Given Data:
1. Mass of Copper: 2 pounds
- Convert pounds to grams: [tex]\(2 \text{ pounds} \times 454 \text{ grams/pound} = 908 \text{ grams}\)[/tex]
2. Specific Heat of Copper: [tex]\(0.385 \text{ J/(g°C)}\)[/tex]
3. Initial Temperature of Copper (T_initial,Cu): 8°C
4. Final Temperature of System (T_final): 52°C
5. Initial Temperature of Water (T_initial,H2O): 60°C
6. Specific Heat of Water: [tex]\(4.186 \text{ J/(g°C)}\)[/tex]
### Step-by-Step Solution:
1. Calculate the Heat Energy Gained by Copper:
The heat energy (Q) gained or lost is given by the formula:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where
- [tex]\( m \)[/tex] is the mass
- [tex]\( c \)[/tex] is the specific heat capacity
- [tex]\( \Delta T \)[/tex] is the change in temperature
For copper:
- Mass ([tex]\( m_{Cu} \)[/tex]): 908 grams
- Change in temperature ([tex]\( \Delta T_{Cu} \)[/tex]): [tex]\( 52°C - 8°C = 44°C \)[/tex]
[tex]\[ Q_{Cu} = m_{Cu} \cdot c_{Cu} \cdot \Delta T_{Cu} \][/tex]
[tex]\[ Q_{Cu} = 908 \text{ g} \times 0.385 \text{ J/(g°C)} \times 44°C \][/tex]
[tex]\[ Q_{Cu} = 15409.36 \text{ J} \][/tex]
2. Set Up the Heat Lost by Water:
For water to cool from 60°C to 52°C, the heat lost can be calculated using the same formula. Assume the mass of water is [tex]\( m_{H2O} \)[/tex] grams (to find):
- Change in temperature ([tex]\( \Delta T_{H2O} \)[/tex]): [tex]\( 60°C - 52°C = 8°C \)[/tex]
[tex]\[ Q_{H2O} = m_{H2O} \cdot c_{H2O} \cdot \Delta T_{H2O} \][/tex]
[tex]\[ Q_{H2O} = m_{H2O} \cdot 4.186 \text{ J/(g°C)} \cdot 8°C \][/tex]
3. Equate the Heat Lost by Water to the Heat Gained by Copper:
[tex]\[ Q_{H2O} = Q_{Cu} \][/tex]
[tex]\[ m_{H2O} \cdot 4.186 \text{ J/(g°C)} \cdot 8°C = 15409.36 \text{ J} \][/tex]
4. Solve for the Mass of Water:
[tex]\[ m_{H2O} \cdot 4.186 \text{ J/(g°C)} \cdot 8°C = 15409.36 \text{ J} \][/tex]
[tex]\[ m_{H2O} \cdot 33.488 = 15409.36 \][/tex]
[tex]\[ m_{H2O} = \frac{15409.36}{33.488} \][/tex]
[tex]\[ m_{H2O} \approx 460.26 \text{ grams} \][/tex]
Since 1 milliliter (mL) of water has a mass of 1 gram, the volume of water needed is the same as its mass:
[tex]\[ \text{Volume of water} = 460.26 \text{ mL} \][/tex]
### Answer:
You would need approximately 460.26 milliliters of water at 60°C to get a final temperature of 52°C of the system after dropping in a 2-pound piece of copper that was at 8°C.
