Answer:
t = 1130 s = 18.83 min
Explanation:
First, we will calculate the energy required to evaporate 500 g of water:
[tex]E = mL[/tex]
where,
E = Energy Required for evaporation of water =?
m = mass of water = 500 g = 0.5 kg
L = Latent heat of vaporization of water = 2.26 MJ/kg = 2260 KJ/kg
Therefore,
[tex]E = (0.5\ kg)(2260\ KJ/kg)\\E = 1130\ KJ[/tex]
Now, we will calculate the time required:
[tex]P = \frac{W}{t}\\\\t = \frac{W}{P}[/tex]
where,
t = time = ?
P = Power of kettle = 1 KW
Therefore,
[tex]t = \frac{1130\ KJ}{1 KW}\\\\[/tex]
t = 1130 s = 18.83 min
