Question

Water is boiled in a 1-L coffee maker with an immersion-type electric heating element at atmospheric

pressure. It was observed that ½-L had evaporated in 13 minutes from the boiling start.
a) If the heating element is rated based on the power required to evaporate water, determine the
rating of the heating element.
b) For the same heating element, how long will it take to raise the 1-L temperature from 18 oC?
C to the boiling point at the same atmospheric pressure?
Take liquid water density (rwater) = 1000 kg/m3 and specific heat (cpwater) = 4.18 kJ/kg.
K).

Answer 1

The rating of the heating element that evaporated 0.5 L of water in 13 minutes is approximately 1,442 W. To raise the temperature of 1 L of water from 18°C to the boiling point with the same heating element, it would take approximately 238 seconds or 3 minutes and 58 seconds.

Step-by-step explanation:

To determine the rating of the heating element based on the power required to evaporate water, we use the energy required to evaporate 1 gram of water which is 2,250 J. Evaporating 0.5 L (500 g) of water in 13 minutes (780 seconds) requires:

Energy = mass × heat of vaporization = 500 g × 2,250 J/g = 1,125,000 J

Power = Energy / Time = 1,125,000 J / 780 s = approximately 1,442 W

Time to Raise the Temperature to Boiling Point

Using the liquid water's specific heat (cp) of 4.18 kJ/kg·K, the mass (m) of the 1-L water which is 1 kg, and the temperature change (ΔT) from 18°C to 100°C (which is 82 K):

Energy required (Q) = m × cp × ΔT = 1 kg × 4.18 kJ/kg·K × 82 K = 343.16 kJ

Since 1 kJ = 1,000 J, the total energy required in joules is 343,160 J. With the power of 1,442 W, the time (t) to heat the water can be calculated using:

t = Q / Power = 343,160 J / 1,442 W = approximately 238 seconds or 3 minutes and 58 seconds.