If your additional 1.12 gallons of 70F water can be considered free of heating time and cost, then we just need to compare the different quantities of water being heated and the temperature they are raised through:
- 2.82 gallons from 70F to 212F = 2.82 * (212-70) = 400.44 galF
- 4 gallons from 70F to 170F = 4 * (170-100) = 400 galF
(I'm using non-standard units - each unit is the amount of energy requried to raise a gallon one degree F.)
In terms of the energy required to heat the water, they are both equivalent requiring 400 units of energy, so they are in principle equal in efficiency and in time taken. Now, I would say that the second approach, heating 4 gallons to 170 is more efficient:
the 1.12 gallons you used in the first example, probably isn't free, adding to the energy needed by the first approach.
higher water temperatures have higher thermal losses, since there is a larger difference between the water temperature and the environment. Hence, the boiling water will loose more heat each second than the 170F water, requiring more energy, and more time if the same sized flame is used as when heating 4 gallons.
There is one counter-argument that favors the first approach on a practical level in terms of time:
- if you don't have temperature control on the propane burner, then you may end up reducing the flame to avoid overshooting the 170F target. When heating to boiling, you can turn the burner all the way up and still not overshoot. You will then probably use more energy, since bigger flames are usually less efficient, but it will heat quicker.
To sum up, with the same-sized flame, the 2nd approach, heating all 4 gallons, is the most energy efficient and quickest.