So the entire heating process from start to all the water boiled away can be broken down into two steps.
Step 1: heating the water from 30C to 100C.
Let's use this formula
Q1 = (mcΔT)w + (mcΔT)Al
Since ΔT is the same for aluminum and water in this step (they're both increasing from 30 to 100), we can factor ΔT out.
Q1 = ((mc)w + (mc)Al)*ΔT
Where,
Q1: heat required to heat the system from 30 to 100
(mc)w: mass (m), heat constant (c), and temperature change (ΔT) for water
(mc)Al: mass (m), heat constant (c), and temperature change (ΔT) for aluminum
ΔT: temperature change
For water:
m = 2.2 kg
c = 4186 J/kgC (this is a constant specific to water, you need to look this up in a reference table)
For aluminum:
m = 0.75 kg
c = 900 J/kgC (this is the heat constant for aluminum)
ΔT = 100-30 = 70 C
Now let's plug in the variables we know:
Q1 = (2.2*4186 + 0.75*900)*70 = 691894 J
Step 2: heating the boiling water enough to vaporize it all
Let's use this formula:
Q2 = (mVap)*(LVap)
Where,
Q2: heat required to boil all the water away
mVap: mass of the system after the water boils away, so just the mass of the empty pot
LVap: latent heat of vaporization of water (this is a constant for water that you need to look up)
mVap = 0.75 kg
LVap = 2256000 J/kg
Now let's plug in the variables we know:
Q2 = 0.75*2256000 = 1692000 J
So now that we have the heat required for both steps, we can simply add those to get the total heat required.
QTotal = Q1 + Q2 = 691894 + 1692000
QTotal = 2383894 J
^Answer for part A
Power is the rate at which heat is transferred. 550W = 550J/s.
At 550 J/s, we can find how long it takes for 2383894 J to transfer.
P = Q/t
Where,
P: power, Q: total heat, t: time required (in seconds)
550 = 2383894/t
t = 2383894/550
t = 4334.35 s = 72.24 min
^Answer for part B