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To solve the problem of determining the final temperature when a piece of iron is placed in water, we will use the principle of conservation of energy. Specifically, the heat lost by the iron will be equal to the heat gained by the water, since we assume no heat is lost to the cup or surroundings.
Here's a step-by-step solution:
1. Convert Initial Temperatures from Kelvin to Celsius:
- Initial temperature of iron: [tex]\( 368 \, K - 273.15 = 94.85 \, ^{\circ}C \)[/tex]
- Initial temperature of water: [tex]\( 268 \, K - 273.15 = -5.15 \, ^{\circ}C \)[/tex]
2. Assign Given Values:
- Mass of iron ([tex]\( m_{iron} \)[/tex]): [tex]\( 25.8 \, g \)[/tex]
- Mass of water ([tex]\( m_{water} \)[/tex]): [tex]\( 25.8 \, g \)[/tex] (since 1 mL of water is approximately 1 g)
- Specific heat capacity of iron ([tex]\( c_{iron} \)[/tex]): [tex]\( 0.449 \, J/g \, ^{\circ}C \)[/tex]
- Specific heat capacity of water ([tex]\( c_{water} \)[/tex]): [tex]\( 4.18 \, J/g \, ^{\circ}C \)[/tex]
- Initial temperature of iron ([tex]\( T_{i_{iron}} \)[/tex]): [tex]\( 94.85 \, ^{\circ}C \)[/tex]
- Initial temperature of water ([tex]\( T_{i_{water}} \)[/tex]): [tex]\( -5.15 \, ^{\circ}C \)[/tex]
3. Set up the Heat Transfer Equation:
- Heat lost by iron ([tex]\( Q_{iron} \)[/tex]): [tex]\( Q_{iron} = m_{iron} \times c_{iron} \times (T_{i_{iron}} - T_{final}) \)[/tex]
- Heat gained by water ([tex]\( Q_{water} \)[/tex]): [tex]\( Q_{water} = m_{water} \times c_{water} \times (T_{final} - T_{i_{water}}) \)[/tex]
4. Equalize the Heat Transfer (since [tex]\( Q_{iron} = -Q_{water} \)[/tex]):
[tex]\[ m_{iron} \times c_{iron} \times (T_{i_{iron}} - T_{final}) = m_{water} \times c_{water} \times (T_{final} - T_{i_{water}}) \][/tex]
5. Insert the Values:
[tex]\[ 25.8 \times 0.449 \times (94.85 - T_{final}) = 25.8 \times 4.18 \times (T_{final} - (-5.15)) \][/tex]
6. Simplify the Equation (cancel out the masses since they are the same):
[tex]\[ 0.449 \times (94.85 - T_{final}) = 4.18 \times (T_{final} + 5.15) \][/tex]
7. Expand and Reorganize the Equation:
[tex]\[ 0.449 \times 94.85 - 0.449 \times T_{final} = 4.18 \times T_{final} + 4.18 \times 5.15 \][/tex]
[tex]\[ 42.58965 - 0.449 \times T_{final} = 4.18 \times T_{final} + 21.557 \][/tex]
8. Combine Like Terms:
[tex]\[ 42.58965 - 21.557 = 4.18 \times T_{final} + 0.449 \times T_{final} \][/tex]
[tex]\[ 21.03265 = 4.629 \times T_{final} \][/tex]
9. Solve for [tex]\( T_{final} \)[/tex]:
[tex]\[ T_{final} = \frac{21.03265}{4.629} \approx 4.55 \, ^{\circ}C \][/tex]
10. Convert Final Temperature Back to Kelvin:
[tex]\[ T_{final} \, (K) = 4.55 \, ^{\circ}C + 273.15 \approx 277.70 \, K \][/tex]
Therefore, the final temperature of the water and the system is approximately 277.70 K.
The correct answer from the list can be determined by checking the closest option, which in this case should be none of them if it was a multiple-choice question with given options. If you were provided the same numeric answer list, the right answer would be right here calculated.
Here's a step-by-step solution:
1. Convert Initial Temperatures from Kelvin to Celsius:
- Initial temperature of iron: [tex]\( 368 \, K - 273.15 = 94.85 \, ^{\circ}C \)[/tex]
- Initial temperature of water: [tex]\( 268 \, K - 273.15 = -5.15 \, ^{\circ}C \)[/tex]
2. Assign Given Values:
- Mass of iron ([tex]\( m_{iron} \)[/tex]): [tex]\( 25.8 \, g \)[/tex]
- Mass of water ([tex]\( m_{water} \)[/tex]): [tex]\( 25.8 \, g \)[/tex] (since 1 mL of water is approximately 1 g)
- Specific heat capacity of iron ([tex]\( c_{iron} \)[/tex]): [tex]\( 0.449 \, J/g \, ^{\circ}C \)[/tex]
- Specific heat capacity of water ([tex]\( c_{water} \)[/tex]): [tex]\( 4.18 \, J/g \, ^{\circ}C \)[/tex]
- Initial temperature of iron ([tex]\( T_{i_{iron}} \)[/tex]): [tex]\( 94.85 \, ^{\circ}C \)[/tex]
- Initial temperature of water ([tex]\( T_{i_{water}} \)[/tex]): [tex]\( -5.15 \, ^{\circ}C \)[/tex]
3. Set up the Heat Transfer Equation:
- Heat lost by iron ([tex]\( Q_{iron} \)[/tex]): [tex]\( Q_{iron} = m_{iron} \times c_{iron} \times (T_{i_{iron}} - T_{final}) \)[/tex]
- Heat gained by water ([tex]\( Q_{water} \)[/tex]): [tex]\( Q_{water} = m_{water} \times c_{water} \times (T_{final} - T_{i_{water}}) \)[/tex]
4. Equalize the Heat Transfer (since [tex]\( Q_{iron} = -Q_{water} \)[/tex]):
[tex]\[ m_{iron} \times c_{iron} \times (T_{i_{iron}} - T_{final}) = m_{water} \times c_{water} \times (T_{final} - T_{i_{water}}) \][/tex]
5. Insert the Values:
[tex]\[ 25.8 \times 0.449 \times (94.85 - T_{final}) = 25.8 \times 4.18 \times (T_{final} - (-5.15)) \][/tex]
6. Simplify the Equation (cancel out the masses since they are the same):
[tex]\[ 0.449 \times (94.85 - T_{final}) = 4.18 \times (T_{final} + 5.15) \][/tex]
7. Expand and Reorganize the Equation:
[tex]\[ 0.449 \times 94.85 - 0.449 \times T_{final} = 4.18 \times T_{final} + 4.18 \times 5.15 \][/tex]
[tex]\[ 42.58965 - 0.449 \times T_{final} = 4.18 \times T_{final} + 21.557 \][/tex]
8. Combine Like Terms:
[tex]\[ 42.58965 - 21.557 = 4.18 \times T_{final} + 0.449 \times T_{final} \][/tex]
[tex]\[ 21.03265 = 4.629 \times T_{final} \][/tex]
9. Solve for [tex]\( T_{final} \)[/tex]:
[tex]\[ T_{final} = \frac{21.03265}{4.629} \approx 4.55 \, ^{\circ}C \][/tex]
10. Convert Final Temperature Back to Kelvin:
[tex]\[ T_{final} \, (K) = 4.55 \, ^{\circ}C + 273.15 \approx 277.70 \, K \][/tex]
Therefore, the final temperature of the water and the system is approximately 277.70 K.
The correct answer from the list can be determined by checking the closest option, which in this case should be none of them if it was a multiple-choice question with given options. If you were provided the same numeric answer list, the right answer would be right here calculated.
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