# GCSE Physics : Temperature & Gravitational potential energy

• BTEC Michio Kaku
The question is how far does it move. Try plotting it. In summary, a student rotates a tube filled with lead shots 50 times, causing a total decrease in the gravitational potential energy of 24.5J. Based on the Specific Heat Capacity of lead shots (160J/kg°C), the rise in temperature after 50 rotations can be calculated. However, the actual temperature rise is likely to be less than this calculated value due to factors such as friction and air resistance. It is also important to note that the net change in gravitational potential energy is zero, as the lead shots fall back to their original position after each rotation. The exact amount of GPE lost in each rotation would depend on the specific height and gravitational field strengthf

## 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.

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.

A student rotates the tube 50 times.
It is unclear whether a rotation is 180 degrees or 360.
Judging from this:
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.
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.

the second set of values is the final GPE
What second set of values? Why subtract one rotation?
for the other two questions I could not think of anything.
What happened to all that lost GPE?
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.