1) On the construction site an electric crane is used to lift materials to the 10th floor. A 180 kW motor requires 2.0 minutes to lift a 10500 kg load if bricks 218 m. How efficient is the crane?
2. Burning 1.0 kg of coal releases 33 MJ of energy. This energy is converted into electricity with an efficiency of 18%.Determine the mass of coal that must be used to supply a city with 2.0MWh.
3. It takes a natural gas heater (methane) a considerable amount of time to heat a swimming pool. A 150 kW heater takes 28 hours to heat 110000L (110000kg) of water by 8.00 degrees Celsius. How efficient is it?
No idea how to solve these problems. I know the efficiency formula but have no idea how to go from there.
electrical work in = 180,000 *120 = 21.6 *10^6 Joules
mechanical work out = 10,500*9.81*218 = 22.5 *10^6 Joules
magic? on moon?
let's see ? 10th floor about 9 * 3 meters high or 27 meters, not 218 meters
2.
1 watt hr = 3600 watt seconds or Joules
so
2.0 Mwh = 7200 MJ
33 MJ/kg * m in kg *.18 = 7200 MJ
m = 1212 kg
If you understand 2, I think you can do 3
You need to look up heat capacity of water in Joules/kg deg C
I got 12.88% for the efficiency if I used 27 m for the height. If I use 218 for the height I got 96.1%
I understand how you did 2 but I have doubts about my answer for 3
I got 1.95% for efficiency...
water is 4186 Joules/kg deg C
110,000 kg
8 deg C
so heat out into water = 4186*110,000*8
= 3.68 * 10^9 Joules into water
electricity in = 150,000 * 28 * 3600
= 1.51*10^10 Joules
efficiency = out/in = .368/1.51 = .244
= 24.4 %
Does the heat capacity vary? My book said it is 4181 J
LOL, yes, depends on purity of water etc. Those numbers are the same to me :)
Thank you!
You are welcome.
To solve these problems, you can use the concept of efficiency, which is the ratio of useful output to the input. In these scenarios, the input is either energy or power, and the output is the work done or the energy released. The formula for efficiency is:
Efficiency = (Useful output / Input) x 100%
Now, let's go through each problem step by step:
1) To determine the efficiency of the electric crane, we need to find the useful output and the input. In this case, the useful output is the work done to lift the load, and the input is the electrical power required by the motor.
- First, we need to find the work done. The formula for work is: Work = Force x Distance. In this case, the force is the weight of the load, which can be calculated by multiplying the mass (10500 kg) by the gravitational acceleration (9.8 m/s^2). So, the force is 10500 kg x 9.8 m/s^2 = 102900 N.
- The distance is given as 218 m.
Now, we can calculate the work done: Work = Force x Distance = 102900 N x 218 m = 22,455,200 Nm.
- The input is given as 180 kW, which is equal to 180,000 W.
Finally, we can calculate the efficiency: Efficiency = (Useful output / Input) x 100%
Efficiency = (22,455,200 Nm / 180,000 W) x 100% = 12475.56%
So, the efficiency of the crane is approximately 12475.56%.
2) To determine the mass of coal required to supply a city with a certain amount of electricity, we need to calculate the energy output and the input.
- The energy output is given as 2.0 MWh (megawatt-hours), which can be converted to joules using the conversion: 1 MWh = 3.6x10^9 J. So, the energy output is 2.0 MWh x 3.6x10^9 J/MWh = 7.2x10^9 J.
- The input energy is calculated by dividing the energy output by the efficiency, which is 18%. So, the input energy is: Input energy = Energy output / Efficiency = 7.2x10^9 J / 0.18 = 4x10^10 J.
- Now, we know that burning 1.0 kg of coal releases 33 MJ (megajoules) of energy, which can be converted to joules using the conversion: 1 MJ = 1x10^6 J. So, the energy released per kg of coal is 33 MJ x 1x10^6 J/MJ = 3.3x10^7 J/kg.
To determine the mass of coal required, we can divide the input energy by the energy released per kg of coal: Mass of coal = Input energy / Energy released per kg.
Mass of coal = 4x10^10 J / 3.3x10^7 J/kg = approximately 1209.09 kg.
So, approximately 1209.09 kg of coal must be used to supply the city with 2.0 MWh of electricity.
3) To determine the efficiency of the natural gas heater, we need to find the useful output and the input.
- The useful output is the amount of energy transferred to the water, which can be calculated using the specific heat capacity of water (approximately 4.18 J/g°C) and the mass of water heated (110000 kg) by the temperature rise (8°C). So, the useful output is: Useful output = Specific heat capacity x Mass x Temperature rise = 4.18 J/g°C x 110000 kg x 8°C = 3.65x10^10 J.
- The input is given as 150 kW, which is equal to 150,000 W.
Now, we can calculate the efficiency: Efficiency = (Useful output / Input) x 100%
Efficiency = (3.65x10^10 J / 150,000 W) x 100% = 2.43x10^7 %
So, the efficiency of the natural gas heater is approximately 2.43x10^7 %.