A power station with an efficiency of 0.4 generates 10(to the power of 8)watts of electric power and dissipates 1.5 x 10(to the power of 8) watts of thermal energy to the cooling water that flows through it. Knowing that the specific heat of water in SI units is 4,184j/kg degrees clecius. Calculate how many kg of water flows through the plant each second if the water is heated through 3 degrees celcius.
(mass flow rate)*C*(delta T) = 1.5*10^8 W
C is the specific heat of the cooling water.
"delta T" is the change in temperature of the water (3.0 C)
Solve for the mass flow rate. It will be in kg/s if you use the specific heat, temperature and power units above.
Why did the power station join a swim team? Because it wanted to make a splash with its cooling water!
Now, let's calculate the amount of water flowing through the power station each second. To do this, we'll use the principle of conservation of energy.
The thermal energy dissipated by the power station is equal to the heat absorbed by the water flowing through it. The formula for heat energy is given by:
Q = m * c * ΔT,
where Q is the thermal energy, m is the mass of the water, c is the specific heat of water, and ΔT is the temperature change.
We already know that Q is equal to 1.5 x 10^8 watts, ΔT is 3 degrees Celsius, and c is 4,184 J/kg degrees Celsius.
By rearranging the formula, we can solve for m:
m = Q / (c * ΔT).
m = (1.5 x 10^8) / (4,184 * 3).
m ≈ 11,298 kg.
So, approximately 11,298 kg of water flow through the power station each second. That's a whole lot of water! Please keep in mind that this calculation is a simplified estimate and may not account for all variables in a real-world scenario.
To calculate the mass of water flowing through the power plant each second, we need to use the equation:
Thermal energy = mass × specific heat × temperature change
Given:
Efficiency of the power station (η) = 0.4
Electric power generated (P) = 10^8 watts
Thermal energy dissipated (Q) = 1.5 × 10^8 watts
Specific heat of water (c) = 4,184 J/kg°C
Temperature change (ΔT) = 3°C
We can start by calculating the electric energy generated using the efficiency formula:
Electric energy generated (E) = Efficiency × Input energy
E = η × P
Substituting the values:
E = 0.4 × 10^8
Next, we can calculate the mass of water from the thermal energy equation:
Q = m × c × ΔT
Since the thermal energy dissipated equals the electric energy generated (Q = E), we can substitute the values:
E = Q = m × c × ΔT
Substituting the known values:
0.4 × 10^8 = m × 4184 × 3
Now, we can solve for the mass (m):
m = (0.4 × 10^8) / (4184 × 3)
Calculating this expression will give us the mass of water flowing through the plant each second.
To solve this problem, we need to use the equation:
Power Input = Power Output + Power Dissipated
In this case, the power input is the electric power generated by the power station, which is 10^8 watts. The power output is the thermal energy dissipated to the cooling water, which is 1.5 x 10^8 watts. As mentioned, the efficiency of the power station is 0.4
We can use these values to find the amount of thermal energy dissipated to the cooling water:
Power Dissipated = Efficiency * Power Input
1.5 x 10^8 watts = 0.4 * 10^8 watts
Now, we can calculate the amount of thermal energy absorbed by the water:
Q = mcΔT
Where Q is the thermal energy absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
We know that the thermal energy dissipated is equal to the thermal energy absorbed:
1.5 x 10^8 watts = mcΔT
Substituting the specific heat capacity of water (c = 4184 J/kg°C) and the change in temperature (ΔT = 3°C) into the equation:
1.5 x 10^8 watts = m * 4184 J/kg°C * 3°C
Simplifying further:
1.5 x 10^8 watts = m * 12552 J/kg
Now, we can solve for the mass of water (m):
m = (1.5 x 10^8 watts) / (12552 J/kg)
m ≈ 11941.3 kg
Therefore, approximately 11941.3 kg of water flows through the power plant each second if the water is heated through 3 degrees Celsius.