How to Generate Heat with Joule Effect?
If you study physics you should learn that

$$ t = \frac {\Delta T \times m \times SHC}{P} $$

where P is power in watts (W), t is time in seconds (s), SHC is specific heat capacity (kJ/kgK) and m is mass (kg). SHC for water is 4.2 kJ/kgK.

Testing for a 2.1 kW kettle with one litre of water at room temperature and rearranging the formula we get

$$ t = \frac {\Delta T \times m \times SHC}{P} = \frac {(100 - 20) \times 1 \times 4.2}{2.1} = 80 \times 2 = 160 \; s$$
So a bit under three minutes to boil a litre of water from room temperature. This sounds about right.

Since you want to know what power to use you should rearrange the formula:

$$ P = \frac {\Delta T \times m \times SHC}{t} $$

Plug in your \$ \Delta T, \; m, SHC \$ and \$ t \$ and calculate the power you think you need.

If you know your supply voltage you can now work out the resistor value from 

$$ R = \frac {V^2}{P} $$

Finally you can work out the current drawn from your battery from

$$ I = \frac {P}{V} $$

All of the above assumes a well insulated container. An open top or conducting sides will cause heat loss.




  The power supply is with a USB cable.


You can probably get 1 A at 5 V from a USB power supply. From \$ P = VI \$ you can see that you have 5 W for your heater. Plug that into the formula for calculating \$ t \$ and see if you think heating coffee from a USB port is feasible.