Final answer:
The time required to melt a 3.0 kg piece of ice at a heat addition rate of 636.0 kW is 1.6 seconds. To calculate this, the heat required to melt the ice (Q = mLf) is equated to the heat energy supplied over time (Pt), resulting in the correct answer being (a) 1.6 s.
Step-by-step explanation:
The student has asked how long it will take for a 3.0 kg piece of ice at 0.0 degrees Celsius to melt when heat is added at a rate of 636.0 kW. To answer this, we must use the formula Q = mLf where Q is the amount of heat, m is the mass of the ice, and Lf is the latent heat of fusion for water.
The heat required to melt 3.0 kg of ice is given by:
Q = (3.0 kg)(334 kJ/kg) = 1002 kJ,
which converts to 1002 x 10^3 J since 1 kJ = 1000 J.
Since power is the rate at which heat is added, the time t required to melt the ice can be found using Pt = Q, where P is power. Substituting the given values into this equation, we get:
636 kW x t = 1002 x 10^3 J
Becuase 1 kW = 1000 J/s, we rewrite 636 kW as 636 x 10^3 J/s. So, t = (1002 x 10^3 J) / (636 x 10^3 J/s)
Calculating t, we get
t = 1.5754717 s ≈ 1.6 s,
So the correct answer is (a) 1.6 s.