A power station with an efficiency of 0.35 generates 10^8 W of electric power and dissipates 1.86 ✕ 10^8 W of thermal energy to the cooling water that flows through it. Knowing the specific heat of water in SI units is 4184 J/kg°C, calculate how many kilograms of water flow through the plant each second if the water is heated through 2 degrees Celsius.
energy to water per second = 1.86*10^8 Joules
Kg water per second heated = m
1.86*10^8 = 4184 * m * 2
To calculate the number of kilograms of water flowing through the power station each second, we need to use the given information about the thermal energy dissipated, the specific heat of water, and the temperature rise.
The formula to calculate thermal energy is:
Thermal energy (Q) = mass (m) * specific heat (c) * temperature change (ΔT)
Given that Q = 1.86 × 10^8 W and ΔT = 2°C, and specific heat c = 4184 J/kg°C, we can rearrange the formula to solve for mass (m).
Q = m * c * ΔT
Rearranging the equation, we have:
m = Q / (c * ΔT)
Substituting the values into the equation:
m = 1.86 × 10^8 W / (4184 J/kg°C * 2°C)
m = 1.86 × 10^8 W / 8368 J/kg
m ≈ 22207.65 kg
Therefore, approximately 22207.65 kilograms of water flow through the power station each second.