Answer:
t = 2.26 x 10⁵ s
Explanation:
The energy supplied to the water will be equal to the heat required for the boiling of water:
E = ΔQ
Pt = mL
where,
P = Power = 10 W
t = time = ?
m = mass of water = 1 kg
L = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
[tex](10\ W)t = (1\ kg)(2.26\ x\ 10^6\ J/kg)\\\\t = \frac{2.26\ x\ 10^6\ J}{10\ W}\\\\[/tex]
t = 2.26 x 10⁵ s
This time will be less than the actual time taken due to some heat loss during the transmission of this heat energy to the container in which water is held.
