A coffee maker can be modeled as a heating element (resistance R) connected to the outlet voltage of 120.0 V (assumed to be dc). The heating element boils small amounts of water at a time as it brews the coffee. When bubbles of water vapor form, they carry liquid water up through the tubing. Because of this, the coffee maker boils 2.10% of the water that passes through it; the rest is heated to 100°C but remains liquid. Starting with water at 10.0°C, the coffee maker can brew 1.00 L of coffee in 8.60 min. Find the resistance R.

To solve this situation, we will need to calculate the amount of energy necessary to elevate the water’s temperature from 10°C to 100°C; then we will calculate the amount of energy necessary to evaporate 2,10% of the water. Let us remember than during the phase change, the water’s temperature remains constant. It is also good to keep in mind that 1 L of water weights 1 Kg exactly

The energy necessary to raise the temperature is:

Q = m*c*∆T = 1 kg* 4186 J/(kg °C) * (100°C-10°C) = 376740 J

Additional to that, the energy required to evaporate 2,1% of this amount of water is:

Q_v = m*L_v = 0,021 * 2257 kJ/kg = 47,397 kJ = 47397 J

Now then, the sum of these two components will give us the amount of energy necessary to achieve the desired state. It is stated that the coffee machine brews the coffee in 8,6 minutes; we can calculate the power necessary to achieve this process by applying the following equation:

[tex]Pow = Q_tot/t = (376740 J +47397 J)/8,6 min = 424137 J/8,6 min * 1 min/ 60 s =821,97 J/s = 821,97 W[/tex]

This is the power that must be supplied by the coffee maker to complete the process. Let us assume that all the electrical power is transformed into heat and that there are no losses of any kind in the system. Remembering the equation of electrical power, we have that:

Pow = V^2/R → R = V^2/Pow = (120.0 V)^2/(821,97 w) = 17,52 Ω

I hope everything was clear with the explanation. If I can help you with anything else, let me know. Have a nice day :D