You are using a gas-powered pump to raise 100 kg of water to a height of 100 m. The energy efficiency of the pump is 20%. By what amount has the internal energy of the water and pump changed?

Given:
- Change in potential energy (\( \Delta PE \)) = \( mgh \)
- \( \Delta PE = 100 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 100 \, \text{m} \)
- \( \Delta PE = 98.1 \, \text{kJ} \)

However, we need to account for the efficiency to find the total energy input.

The energy input (\( E_{input} \)) is related to the efficiency (\( \eta \)) and the change in potential energy:
\[ \eta = \frac{\Delta PE}{E_{input}} \]

Rearranging for \( E_{input} \):
\[ E_{input} = \frac{\Delta PE}{\eta} \]
\[ E_{input} = \frac{98.1 \, \text{kJ}}{0.20} \]
\[ E_{input} = 490.5 \, \text{kJ} \]

Now, the increase in internal energy is the difference between the energy input and the change in potential energy:
\[ \Delta E_{internal} = E_{input} - \Delta PE \]
\[ \Delta E_{internal} = 490.5 \, \text{kJ} - 98.1 \, \text{kJ} \]
\[ \Delta E_{internal} = 392.4 \, \text{kJ} \]

Therefore, the internal energy of the water and pump has increased by 392.4 kJ.