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.

An energy efficiency of 20% means that, of 100 % of energy that the pump obtains from the gas, uses 20% of that energy to perform work over the water, and losses the other 80% in the form of heat.

This means that the change in potential energy you obtained is 20% of the total internal energy that the pump has consumed:

[tex]0.2 * \Delta E_{pump_{internal}} = - \Delta E_{potential} = [/tex]

[tex] \Delta E_{pump_{internal}} = - \frac{1}{0.2} * \Delta E_{potential}[/tex]

[tex] \Delta E_{pump_{internal}} = - 5 * \Delta E_{potential}[/tex]

[tex] \Delta E_{pump_{internal}} = - 5 * 98.1 \ kJ[/tex]

[tex] \Delta E_{pump_{internal}} = - 490.5 \ kJ[/tex]

The water is raised, changing its potential energy, but, this doesn't contributes to its internal energy. But the heat that was loosed by the pump does. Using the approximation, as good physicist, that the pump doesn't exchange heat with the surroundings, and only transfers it to the water. The water gains the heat loosed by the pump.

[tex] \Delta E_{water_{internal}} = - 0.8 * \Delta E_{pump_{internal}} = - 0.8 * 490.5 \ kJ[/tex]

[tex] \Delta E_{water_{internal}} = 392.4\ kJ[/tex]