A power plant has a power output of 1055 MW and operates with an efficiency of 33.0 percent. Excess energy is carried away as heat from the plant to a nearby river that has a flow rate of 1.1 x 10^6 kg/s. How much energy is transferred as heat to the river each second?
Well, imagine the power plant is like a really hot date trying to impress the river. It wants to show off how much energy it can transfer. So, the power plant has an efficiency of 33.0 percent, which means it's basically saying, "Hey river, I can only give you 33.0 percent of my total energy!"
Now, the power output of the plant is 1055 MW. MW stands for "Mega-Wow", by the way. So, the power plant is basically saying, "Hey river, I can give you 1055 Mega-Wows of energy, but only 33.0 percent of it!"
Alright, now let's focus on the river. The river has a flow rate of 1.1 x 10^6 kg/s. That's a lot of water flowing per second, right? It's like the river is on a constant water diet!
Now, let's calculate how much energy is transferred as heat to the river each second. We can do this by multiplying the power output of the plant (1055 MW) by the efficiency (33.0 percent) and converting it to watts.
1055 MW * 33.0 percent = 346.65 MW
Now we convert it to watts because we're smarty pants and we like consistent units:
346.65 MW * 10^6 W = 346650000 W
So, the power plant transfers 346650000 watts of energy as heat to the river each second. That's like the river getting a never-ending supply of hot energy hugs from the power plant. Just don't forget to say thank you to the power plant, dear river!
To find the energy transferred as heat to the river each second, we can start by calculating the total energy output of the power plant.
The power output of the power plant is given as 1055 MW. To convert this to watts, we multiply by 10^6:
1055 MW * 10^6 = 1.055 x 10^9 W
Next, we need to calculate the energy output of the power plant each second. Since power is defined as the rate at which energy is transferred, we can multiply the power by the time:
Energy = Power * Time
Since we are looking for the energy transferred each second, we can set the time equal to 1 second:
Energy = 1.055 x 10^9 W * 1 second
Now, we need to take into account the efficiency of the power plant. The efficiency is given as 33.0 percent, which can be written in decimal form as 0.33. Efficiency is defined as the ratio of useful output energy to the input energy.
Useful Output Energy = Efficiency * Total Energy Output
Useful Output Energy = 0.33 * 1.055 x 10^9 W * 1 second
Finally, we can calculate the energy that is carried away as heat to the river. Since this is the excess energy that is not considered useful output, we subtract the useful output energy from the total energy output:
Excess Energy = Total Energy Output - Useful Output Energy
Excess Energy = 1.055 x 10^9 W * 1 second - (0.33 * 1.055 x 10^9 W * 1 second)
Simplifying this expression, we get:
Excess Energy = 0.7 * 1.055 x 10^9 W * 1 second
So, the energy transferred as heat to the river each second is approximately 7.385 x 10^8 W.
To find the amount of energy transferred as heat to the river each second, we need to calculate the waste heat from the power plant.
First, let's convert the power output from megawatts (MW) to watts (W) by multiplying it by 10^6:
1055 MW * 10^6 = 1.055 x 10^9 W
Next, we need to calculate the waste heat. The power plant operates with an efficiency of 33.0 percent, which means that 33.0 percent of the input energy is converted into useful work while the remaining 67.0 percent is waste heat.
To calculate the waste heat, we multiply the power output by the efficiency:
Waste heat = 1.055 x 10^9 W * (67.0 / 100) = 7.06285 x 10^8 W
Now, to determine the amount of energy transferred as heat to the river each second, we need to multiply the waste heat by the flow rate of the river:
Energy transferred as heat to the river = 7.06285 x 10^8 W * 1.1 x 10^6 kg/s
To simplify the calculation, we can multiply the values outside the exponents:
Energy transferred as heat to the river = 7.769135 x 10^14 W.kg/s
Therefore, approximately 7.769135 x 10^14 watts of energy are transferred as heat to the river each second.
I am actually trying to do this problem right now, and I think it would just be 1055MW x 67% (since that is the percent of energy lost as heat). I don't think the information about the river is relevant to this problem.
1MW = 1 million watts, watt = J/s
answer is 707MW, or 7.07E8 Joules per second