Question

Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance heater with 84.5% efficiency supplies heat to the pot. How many minutes will it take to boil 6.03 kg of water?

Answer 1

Answer:

t = 6179.1 s = 102.9 min = 1.7 h

Explanation:

The energy provided by the resistance heater must be equal to the energy required to boil the water:

E = ΔQ

ηPt = mH

where.

η = efficiency = 84.5 % = 0.845

P = Power = 2.61 KW = 2610 W

t = time = ?

m = mass of water = 6.03 kg

H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

$t = \frac{1.362\ x\ 10^7\ J}{2205.45\ W}$

t = 6179.1 s = 102.9 min = 1.7 h