To find the net work done by the machine, let's break down the information provided:

1. Power Consumption: 3500 kWh.
2. Time of Operation: 600 seconds.
3. Wasted Heat Energy: 500,000 J.

Step-by-Step Solution

1. Convert Power Consumption to Joules:
   Since 1 kWh = 3,600,000 J, we can find the total energy consumed. $3500 \text{ kWh} = 3500 \times 3,600,000 = 12,600,000,000 \text{ J}$

2. Calculate Total Energy Consumed in 600 Seconds: Power is energy per unit time, so the power consumption rate per second is: $\text{Power} = \frac{12,600,000,000 \text{ J}}{600 \text{ s}} = 21,000,000 \text{ J/s}$

3. Calculate Net Work Done: The net work done is the total energy consumed minus the wasted heat energy. $\text{Net Work Done} = 12,600,000,000 \text{ J} - 500,000 \text{ J} = 12,599,500,000 \text{ J}$

Therefore, the net work done by the machine is 12,599,500,000 J.

Would you like more details or have any questions?

1. How is power related to energy and time in physics?
2. Why do we convert kWh to Joules in energy calculations?
3. What are the implications of wasted energy in a machine?
4. How does one measure efficiency based on work done and wasted energy?
5. What other factors can affect a machine's energy efficiency?

Tip: Remember, power (in watts) is the rate of energy consumption per second.