By my calculations.

If we assume that we want to heat 10g of popcorn to 150C , then by my calculations:

Using Specific Heat Capacity:

q = mct

c = 4200, t = 130, m = 0.01

Therefore:

q = 4200 * 130 * 0.01 = 5460 J

Now Power is Joules per Second:

p = q / t

p = 2.5 , q = 5460

Therefore in order to raise the temperature of 10g of pop corn to 150C at maximum USB power it would take:

5460 / 2.5 = t = 2,184 Seconds = 36 Minutes

This assumes that heat transfer is 100% efficient and there is no heat loss.

Now providing that the heat input into the system is greater than the heat loss, there should eventually

be some popcorn.

I think the more interesting question is how long it took.

Another thing to note, is that the USB bus does not provide 500mA of power unless requested by the device.

Therefore it is likely in this situation the USB bus is only providing 100mA, IE 0.5W

But that is just my views.