Water boils under the action of a 1150 W heater. The specific heat of vaporization

Water boils under the action of a 1150 W heater. The specific heat of vaporization of water is 2300 kJ / kg. How much water will boil off in 20 minutes?

These tasks: N (power of the used heater) = 1150 W; t (heater operation time) = 20 min (in SI t = 1200 s).

Constants: according to the condition L (specific heat of vaporization of water) = 2300 kJ / kg = 2.3 * 10 ^ 6 J / kg.

We express the mass of water boiled out in 20 minutes from the equality: L * m = Q = A = N * t, whence m = N * t / L.

Let’s calculate: m = 1150 * 1200 / (2.3 * 10 ^ 6) = 0.6 kg (600 g).

Answer: In 20 minutes of heater operation, 600 grams of water will boil off.