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?​

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,

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.

answer: what are you talking about?

explanation:

6 cos 34, 6 sin 34 trig