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A lead bullet specific heat is 0.032 is completely stopped when it strikes a target with velocity 300 the heat generated is equally shared by bullet and target the rise in temperature of bullet will be?
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A lead bullet specific heat is 0.032 is completely stopped when it str...
To find the rise in temperature of a lead bullet when it strikes a target, we will use the principles of energy conservation and specific heat capacity.
Initial Data
- **Specific Heat of Lead (c)**: 0.032 cal/g°C
- **Velocity of Bullet (v)**: 300 m/s
- **Mass of Bullet (m)**: Assume 1 g for simplicity. (You can generalize later)
Energy Calculation
- **Kinetic Energy (KE)**:
The kinetic energy of the bullet before striking the target is given by the formula:
KE = \( \frac{1}{2} mv^2 \)
Substituting the values:
KE = \( \frac{1}{2} \times 1 \times (300)^2 \) = 45000 J (or 10708.4 cal, since 1 J = 0.239 cal)
Heat Transfer
- **Heat Generated (Q)**:
The total heat generated when the bullet comes to a stop will be equal to its initial kinetic energy:
Q = KE = 10708.4 cal
Heat Sharing
- **Heat Distribution**:
The heat generated is equally shared between the bullet and the target:
Q_bullet = Q_target = \( \frac{Q}{2} \) = \( \frac{10708.4}{2} \) = 5354.2 cal
Temperature Rise Calculation
- **Rise in Temperature (ΔT)**:
The rise in temperature of the bullet can be calculated using the formula:
\( \Delta T = \frac{Q_bullet}{m \cdot c} \)
Substituting the values:
\( \Delta T = \frac{5354.2}{1 \cdot 0.032} \) = 167,000 °C
Conclusion
The rise in temperature of the lead bullet when it strikes the target at 300 m/s is approximately 167,000 °C, which indicates a significant increase in temperature due to the conversion of kinetic energy into thermal energy.
Initial Data
- **Specific Heat of Lead (c)**: 0.032 cal/g°C
- **Velocity of Bullet (v)**: 300 m/s
- **Mass of Bullet (m)**: Assume 1 g for simplicity. (You can generalize later)
Energy Calculation
- **Kinetic Energy (KE)**:
The kinetic energy of the bullet before striking the target is given by the formula:
KE = \( \frac{1}{2} mv^2 \)
Substituting the values:
KE = \( \frac{1}{2} \times 1 \times (300)^2 \) = 45000 J (or 10708.4 cal, since 1 J = 0.239 cal)
Heat Transfer
- **Heat Generated (Q)**:
The total heat generated when the bullet comes to a stop will be equal to its initial kinetic energy:
Q = KE = 10708.4 cal
Heat Sharing
- **Heat Distribution**:
The heat generated is equally shared between the bullet and the target:
Q_bullet = Q_target = \( \frac{Q}{2} \) = \( \frac{10708.4}{2} \) = 5354.2 cal
Temperature Rise Calculation
- **Rise in Temperature (ΔT)**:
The rise in temperature of the bullet can be calculated using the formula:
\( \Delta T = \frac{Q_bullet}{m \cdot c} \)
Substituting the values:
\( \Delta T = \frac{5354.2}{1 \cdot 0.032} \) = 167,000 °C
Conclusion
The rise in temperature of the lead bullet when it strikes the target at 300 m/s is approximately 167,000 °C, which indicates a significant increase in temperature due to the conversion of kinetic energy into thermal energy.
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A lead bullet specific heat is 0.032 is completely stopped when it strikes a target with velocity 300 the heat generated is equally shared by bullet and target the rise in temperature of bullet will be?
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A lead bullet specific heat is 0.032 is completely stopped when it strikes a target with velocity 300 the heat generated is equally shared by bullet and target the rise in temperature of bullet will be? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A lead bullet specific heat is 0.032 is completely stopped when it strikes a target with velocity 300 the heat generated is equally shared by bullet and target the rise in temperature of bullet will be? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A lead bullet specific heat is 0.032 is completely stopped when it strikes a target with velocity 300 the heat generated is equally shared by bullet and target the rise in temperature of bullet will be?.
A lead bullet specific heat is 0.032 is completely stopped when it strikes a target with velocity 300 the heat generated is equally shared by bullet and target the rise in temperature of bullet will be? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A lead bullet specific heat is 0.032 is completely stopped when it strikes a target with velocity 300 the heat generated is equally shared by bullet and target the rise in temperature of bullet will be? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A lead bullet specific heat is 0.032 is completely stopped when it strikes a target with velocity 300 the heat generated is equally shared by bullet and target the rise in temperature of bullet will be?.
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