Answer :
Answer:
0.09 kg
Explanation:
The details of boiling liquid are;
The rate of thermal energy absorption by the liquid, P = 450 W
The specific latent heat of vaporization, c = 2.7 × 10⁶ J/kg
The required information = The amount of liquid vaporized in t = 9 minutes
We have;
P = E/t
Where;
P = Power supplied by the heat source = The rate of absorption of thermal energy
E = The energy absorbed
t = The time
The energy absorbed in 9 minutes, E = P × t
∴ E = 450 W × 9 min × 60 s/min = 243,000 J
The heat required to evaporate 1 kg of liquid = The specific latent heat of vaporization, c = 2.7 × 10⁶ Joules
The mass of the liquid evaporated by the energy absorbed, E, m = E/c
∴ m = 243,000 J/(2.7 × 10⁶ J/kg) = 0.09 kg
Therefore, the mass of the liquid vaporized by the energy absorbed at a rate of 450 W, in 9 minutes, m = 0.09 kg.