Question

If 10 W of power is supplied to 1 kg of water at 100℃, how long will it take to for the water to completely boil away? The time calculated is a little less than actual time of boiling in practice. Why?​

Answer 1

t = 2.26 x 10⁵ s

Step-by-step 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,

`(10\ W)t = (1\ kg)(2.26\ x\ 10^6\ J/kg)\\\\t = (2.26\ x\ 10^6\ J)/(10\ W)\\\\`

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.