You show two different wire configurations, one at .74 V and .11 A, and another at .05 V and .10 A. The fact that the current is essentially the same in both cases is an indication that your measurements are correct, since photodiodes (of which a solar cell is an example) act as current sources when shorted.
Now calculate the power dissipated in the wires. .74 V x .11 A equals about 81 mW, while .05 V x .1 A equals 5 mW. The ratio, 16 should be equal to the ratio of the resistances, which is 5 in this case. The error is probably due to resistance measurement error using your meter. I suspect you failed to zero out the probe resistance.
Now. Look at the actual powers. 80 mW should be detectable, but 5 mW - not so much. So, how are you "detecting" your temperatures? I doubt you're using a thermal camera, and any physical thermometer may well be too big a heat sink to allow easy measurement. Assuming your power source is well-isolated, you might try letting the wire run for a few seconds, then touch it to your (dry) lips, which are very sensitive. Note that, if you do this, the temperature will almost immediately seem to cool to neutral temperature - again, heat sink effects. Of course, this requires absolute certainty that your power supply is isolated, and is not remotely a good idea if you don't know what this means.