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A metal ball of mass 0.5 kg falls freely from a height of 10 m and bounces to a height of 5.5 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is?
(Specific heat of metal = 450J/Kg°C )
- a)0.001°C
- b)0.1°C
- c)10°C
- d)1°C
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A metal ball of mass 0.5 kg falls freely from a height of 10 m and bou...
- The Ball was initially at a height 10 meters above the ground and when it bounced from the ground its height was 5.5 meters above the ground.
- So the energy absorbed and raised temperature of ball was the difference between these two potential energies of the ball which was actually a difference between the initial and final height of the ball.
Given Data:
hinitial = initial height = 10 m , m = 0.5 kg
⟹ hfinal = final height = 5.5 m ,
C = 450 J/kg° C
m = 0.5 kg
∴ Q = mg (hinitial - hfinal)
∴ mC⋅(ΔT) = mg (hinitial - hfinal)
∴ 0.5 × 450 × ΔT = 0.5 × 9.81 × (10 - 5.5)
∴ ΔT = 0.196 ° C ≈ 0.2 ° C
⟹ The nearest value in options are 0.1 ° C so we have to choose it.
hinitial = initial height = 10 m , m = 0.5 kg
⟹ hfinal = final height = 5.5 m ,
C = 450 J/kg° C
m = 0.5 kg
∴ Q = mg (hinitial - hfinal)
∴ mC⋅(ΔT) = mg (hinitial - hfinal)
∴ 0.5 × 450 × ΔT = 0.5 × 9.81 × (10 - 5.5)
∴ ΔT = 0.196 ° C ≈ 0.2 ° C
⟹ The nearest value in options are 0.1 ° C so we have to choose it.
Most Upvoted Answer
A metal ball of mass 0.5 kg falls freely from a height of 10 m and bou...
Given data:
- Mass of the metal ball (m) = 0.5 kg
- Initial height (h1) = 10 m
- Final height after bounce (h2) = 5.5 m
- Specific heat of metal (c) = 450 J/kg°C
Calculating Potential Energy:
The potential energy at height h is given by the formula: PE = mgh
Initial potential energy (PE1) = 0.5 * 9.81 * 10 = 49.05 J
Final potential energy after bounce (PE2) = 0.5 * 9.81 * 5.5 = 27.03 J
Calculating Energy Dissipated:
The energy dissipated during the bounce is the difference in potential energy:
Energy dissipated = PE1 - PE2 = 49.05 - 27.03 = 22.02 J
Calculating Rise in Temperature:
The energy dissipated is absorbed by the metal ball, leading to a rise in temperature. This change in energy is converted to heat, given by the formula: Q = mcΔT
Where Q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Substitute the values:
22.02 = 0.5 * 450 * ΔT
ΔT = 22.02 / (0.5 * 450) = 0.0978 °C
Therefore, the rise in temperature of the metal ball is approximately 0.1 °C. Hence, the correct answer is option B.
- Mass of the metal ball (m) = 0.5 kg
- Initial height (h1) = 10 m
- Final height after bounce (h2) = 5.5 m
- Specific heat of metal (c) = 450 J/kg°C
Calculating Potential Energy:
The potential energy at height h is given by the formula: PE = mgh
Initial potential energy (PE1) = 0.5 * 9.81 * 10 = 49.05 J
Final potential energy after bounce (PE2) = 0.5 * 9.81 * 5.5 = 27.03 J
Calculating Energy Dissipated:
The energy dissipated during the bounce is the difference in potential energy:
Energy dissipated = PE1 - PE2 = 49.05 - 27.03 = 22.02 J
Calculating Rise in Temperature:
The energy dissipated is absorbed by the metal ball, leading to a rise in temperature. This change in energy is converted to heat, given by the formula: Q = mcΔT
Where Q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Substitute the values:
22.02 = 0.5 * 450 * ΔT
ΔT = 22.02 / (0.5 * 450) = 0.0978 °C
Therefore, the rise in temperature of the metal ball is approximately 0.1 °C. Hence, the correct answer is option B.
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A metal ball of mass 0.5 kg falls freely from a height of 10 m and bounces to a height of 5.5 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is?(Specific heat of metal = 450J/Kg°C )a)0.001°Cb)0.1°Cc)10°Cd)1°CCorrect answer is option 'B'. Can you explain this answer?
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A metal ball of mass 0.5 kg falls freely from a height of 10 m and bounces to a height of 5.5 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is?(Specific heat of metal = 450J/Kg°C )a)0.001°Cb)0.1°Cc)10°Cd)1°CCorrect answer is option 'B'. Can you explain this answer? for EmSAT Achieve 2025 is part of EmSAT Achieve preparation. The Question and answers have been prepared according to the EmSAT Achieve exam syllabus. Information about A metal ball of mass 0.5 kg falls freely from a height of 10 m and bounces to a height of 5.5 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is?(Specific heat of metal = 450J/Kg°C )a)0.001°Cb)0.1°Cc)10°Cd)1°CCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for EmSAT Achieve 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metal ball of mass 0.5 kg falls freely from a height of 10 m and bounces to a height of 5.5 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is?(Specific heat of metal = 450J/Kg°C )a)0.001°Cb)0.1°Cc)10°Cd)1°CCorrect answer is option 'B'. Can you explain this answer?.
A metal ball of mass 0.5 kg falls freely from a height of 10 m and bounces to a height of 5.5 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is?(Specific heat of metal = 450J/Kg°C )a)0.001°Cb)0.1°Cc)10°Cd)1°CCorrect answer is option 'B'. Can you explain this answer? for EmSAT Achieve 2025 is part of EmSAT Achieve preparation. The Question and answers have been prepared according to the EmSAT Achieve exam syllabus. Information about A metal ball of mass 0.5 kg falls freely from a height of 10 m and bounces to a height of 5.5 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is?(Specific heat of metal = 450J/Kg°C )a)0.001°Cb)0.1°Cc)10°Cd)1°CCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for EmSAT Achieve 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metal ball of mass 0.5 kg falls freely from a height of 10 m and bounces to a height of 5.5 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is?(Specific heat of metal = 450J/Kg°C )a)0.001°Cb)0.1°Cc)10°Cd)1°CCorrect answer is option 'B'. Can you explain this answer?.
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