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Answer :
The thermal efficiency of the heating process can be determined based on the energy input and the energy output. In this case, the microwave oven consumes energy during standby and heating, and the temperature increase and water evaporation represent the energy output.
To calculate the thermal efficiency, we need to find the total energy input and the useful energy output.
During standby, the oven consumes 0.2 kW for 200 seconds, resulting in an energy input of (0.2 kW) * (200 s) = 40 kJ. During heating, the oven consumes 2.5 kW for 200 seconds, resulting in an energy input of (2.5 kW) * (200 s) = 500 kJ.
The energy output is represented by the temperature increase and water evaporation. The temperature increase from 25°C to 70°C for 1 kg of water corresponds to an energy output of (70°C - 25°C) * (1 kg) * (4.18 kJ/kg°C) = 156.5 kJ. The evaporation of 10 g of water corresponds to an additional energy output of (10 g) * (2260 kJ/kg) = 22.6 kJ.
Therefore, the total useful energy output is 156.5 kJ + 22.6 kJ = 179.1 kJ. The thermal efficiency is calculated by dividing the useful energy output by the total energy input, resulting in a thermal efficiency of (179.1 kJ) / (40 kJ + 500 kJ) = 0.291 or 29.1%.
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