Question

A high-end gas stove usually has at least one burner rated at 14,000 Btu/h.

(a) If you place a 0.37-kg aluminum pot containing 2.2 liters of water at 20°C on this burner, how long will it take to bring the water to a boil, assuming all the heat from the burner goes into the pot?

min

(b) Once boiling begins, how much time is required to boil all the water out of the pot?

min

Answer 1

The total thermal energy going into the pot and the water to bring the water to a boil may be stated as:

Q = Q_pot + Q_water

The time taken to raise the water to boil is:

t = [Q_pot + Q_water]/P

Where P is power in watts, the value of Q is equal to mc?T, so:

t = [mc?T_pot + mc?T_water] / P

Here,

c is specific heat (for aluminum its 900 J/kg?°C and for water, 4186J/kg?°C),

m is mass and

T is temperature.

The mass of the water is 2.2 kg (easy conversion from liters).

The power is found from:

14000Btu/h = 14000BTU/h(0.293W/1.00BTU/h)
= 4100W

part (a):

t = [0.37(900)(80) + 2.2(4186)(80)] / 4100

t = 186.189 s

Hence, it takes 3.1 minutes to bring water to boil.

Part (b)

In this case, the water changes phase, so for water, use the equation Q = mL (where L is the latent heat of vaporization, for water it is 2.26 x 10^6J/kg). :

t = mL_water / P

= (2.2kg)( 2.26 x 10^6J/kg) / 4100W

= 1212.68 s

It takes 20.2 minutes to boil all the water out of the pot.