The mass of the steel pot is 308g (without the lid), and contains 650g of water. The stove was lit, and after the flame settled down the pot was placed on top.
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| The setup |
Each minute the temperature of the pot was measured with an infrared thermometer. It was measured near the corner of the pot from the inside.
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| Temperature vs Time |
The graph from 2 -> 20min is fairly linear, so let's use that part of the graph. At 2min it was 24.8°C, at 20min it was 73.8°C.
ΔTemp = 73.8°C - 24.8°C = 49.0°C
The specific heat capacity of water is 4.18J/g.K, so the amount of energy that's gone into the water is
ΔEwater = 4.18J/g.K * 650g * 49K = 133.1KJ
And the specific heat capacity of steel is 0.47J/g.K, so the amount of energy that's gone into the steel is
ΔEpot = 0.47/g.K * 308g * 49K = 7.1KJ
So the total energy into our system is 133.1KJ + 7.1KJ = 140.2KJ
Now as
Δtime = 20min - 2min = 18min = 1080seconds
our power is 140.2KJ / 1080s = 130W
So it seems we put around 130W into our system, meaning our stove outputs at least that much. How much heat got "lost" ... I've no idea, but it gives us an idea of the sort of numbers we're looking at.
