A lump of lead with mass 0.5kg is dropped from height of 20m onto a hard surface it doesn't rebound but remains there at rest for a long period of time. A, find change in heat , change in work, change in internal energy for the lead during this process?

Answer

give brief explanation, please

To find the changes in heat, work, and internal energy for the lead in this process, we need to consider the different forms of energy involved.

1. Change in heat (ΔQ):
Since the problem states that the lead remains at rest on the hard surface for a long period of time, we can assume that no heat is transferred to or from the lead during the process. Therefore, the change in heat (ΔQ) is zero.

2. Change in work (ΔW):
When the lead is dropped from a height, it undergoes a free-fall motion and eventually comes to a stop when it hits the hard surface. During this process, work is done against the gravitational force. The work done is given by the formula:
ΔW = mgh
where m is the mass of the lead (0.5 kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height (20 m) from which the lead is dropped.

Substituting the values into the formula:
ΔW = 0.5 kg × 9.8 m/s² × 20 m
ΔW = 98 Joules

Therefore, the change in work (ΔW) for the lead is 98 Joules.

3. Change in internal energy (ΔU):
The change in internal energy (ΔU) is equal to the sum of the changes in heat (ΔQ) and work (ΔW). Since we found that ΔQ = 0 and ΔW = 98 Joules, the change in internal energy for the lead is also 98 Joules.

Therefore, the change in internal energy (ΔU) for the lead is 98 Joules.