Question

A high-end gas stove usually has at least one burner rated at 14,000 Btu/h. If you place a 0.47-kg aluminum pot containing 2.3 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?

Please show work, tried problem three times and still didn't get it right.

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 a boil is:

t = Q / P
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 copper its 387J/kg?°C and for water, 4186J/kg?°C), m is mass and T is temperature. The mass of the water is 2.3kg (easy conversion from liters). The power is found from:

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


t = [(0.47kg)( 387J/kg?°C)(80°C) + (2.3kg)( 4186J/kg?°C)(80°C)] / 4100W
= 191.41s (rounded, or 3.2min.)

Hope this helps you.