Answer
First, let’s calculate the amount of thermal energy that is needed to heat the 650 m³ of seawater from 10°C to 100°C:
Q = mcΔT
where Q is the thermal energy, m is the mass of the seawater, c is the specific heat of water (4,184 J/kg·K), and ΔT is the change in temperature.
The mass of 650 m³ of seawater is:
m = ρV
where ρ is the density of seawater (1,025 kg/m³).
m = 1,025 kg/m³ × 650 m³ = 666,250 kg
Now we can calculate Q:
Q = mcΔT = (666,250 kg)(4,184 J/kg·K)(90°C) ≈ 25.1 GJ
The time it takes for 33 MW of thermal power to generate 25.1 GJ of thermal energy is:
t = Q/P
where P is the thermal power.
t = 25.1 GJ/33 MW = 760 seconds ≈ 12.7 minutes
Therefore, it would take approximately 12.7 minutes for the 33 MW of thermal power to bring the 650 m³ of seawater to the boiling point.
Work Step by Step
First, let’s calculate the amount of thermal energy that is needed to heat the 650 m³ of seawater from 10°C to 100°C:
Q = mcΔT
where Q is the thermal energy, m is the mass of the seawater, c is the specific heat of water (4,184 J/kg·K), and ΔT is the change in temperature.
The mass of 650 m³ of seawater is:
m = ρV
where ρ is the density of seawater (1,025 kg/m³).
m = 1,025 kg/m³ × 650 m³ = 666,250 kg
Now we can calculate Q:
Q = mcΔT = (666,250 kg)(4,184 J/kg·K)(90°C) ≈ 25.1 GJ
The time it takes for 33 MW of thermal power to generate 25.1 GJ of thermal energy is:
t = Q/P
where P is the thermal power.
t = 25.1 GJ/33 MW = 760 seconds ≈ 12.7 minutes
Therefore, it would take approximately 12.7 minutes for the 33 MW of thermal power to bring the 650 m³ of seawater to the boiling point.
