GCSE Physics : Temperature & Gravitational potential energy

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SUMMARY

The discussion centers on a GCSE Physics problem involving gravitational potential energy (GPE) and specific heat capacity. The lead shot, weighing 50 grams, experiences a total decrease in GPE of 24.5J after being rotated 50 times in a tube. The specific heat capacity of lead is given as 160J/kg°C, which is used to calculate the temperature rise. However, the temperature rise is expected to be less than calculated due to energy losses not accounted for in the ideal calculations.

PREREQUISITES
  • Understanding of gravitational potential energy calculations
  • Knowledge of specific heat capacity and its formula
  • Familiarity with energy conservation principles
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the concept of energy loss in real-world systems
  • Learn about the implications of specific heat capacity in thermal energy transfer
  • Explore the relationship between mass, height, and gravitational potential energy
  • Investigate the effects of rotational motion on energy distribution
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Students preparing for GCSE Physics exams, educators teaching energy concepts, and anyone interested in the practical applications of thermodynamics and mechanics.

BTEC Michio Kaku
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Homework Statement


Q7. Some lead shot with a mass of 50 grams is placed into a card board box the distance from one end to the other being 1m. The ends are sealed with rubber bungs in order to prevent the lead shots from falling out, the tube is rotated so the lead shots fall down from one end to hit the rubber bung at the opposite end.

A) A student rotates the tube 50 times. Calculate the total decrease in the gravitational potential energy store of the lead shot in the process.

B) The Specific Heat Capacity of the lead shots is 160J/kg°c . Calculate the rise in temperature after the tube has been rotated 50 times.

C) Why is the temperature rise of the lead shot likely to be less than what was calculated in part B.

Homework Equations



Gravitational potential energy = mass x gravitational field strength x height
Change in temperature= Change in energy / mass x specific heat capacity
Final GPE - Initial GPE[/B]

The Attempt at a Solution



Question A ) Firstly, I had calculated the initial GPE of the lead shots which I had derived to be 0.5J ( Note: The value 10 was used for gravitational field strength) as I had done GPE = 0.05kg x 10N/M x 1m. As for the second value because it was said to be flipped 50 times I had multiplied the height by 50, this meant that I did GPE= 0.05kg x 10N/M x 50 = 25J . Assuming the second set of values is the final GPE I did 25J - 0.5J= 24.5J lost

As for the other two questions I could not think of anything.
 
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A) If we turn it 50 times, that is an even number, so maybe I would expect GPE at the end to be the same as at the start?
B) The second equation has everything you need to work out the temp rise when 25J of energy is given to the lead. You know mass and SHC.
 
BTEC Michio Kaku said:
A student rotates the tube 50 times.
It is unclear whether a rotation is 180 degrees or 360.
Judging from this:
BTEC Michio Kaku said:
the tube is rotated so the lead shots fall down from one end to hit the rubber bung at the opposite end.
it probably means 180.
BTEC Michio Kaku said:
total decrease in the gravitational potential energy
This is not clear either. As tech99 points out, the net change in GPE is zero. But you are almost surely right to interpret this as being the sum of all the GPEs lost in each rotation.

BTEC Michio Kaku said:
the second set of values is the final GPE
What second set of values? Why subtract one rotation?
BTEC Michio Kaku said:
for the other two questions I could not think of anything.
What happened to all that lost GPE?
tech99 said:
If we turn it 50 times, that is an even number
Since the shot moves to the end each (half) rotation, it makes no difference whether it is odd or even.
 

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