Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water level in the pan drops by 1.45 cm in 18.6 min, determine the rate of heat transfer to the pan in watts. (Give your answer in 3 significant digits.)

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

Q = 20.22 x 10³ W = 20.22 KW

i don't understand. explain it clearly