Answer:
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
<u>Q = 20.22 x 10³ W = 20.22 KW</u>
