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A power station using coal as fuel has an average power output of 3000 MW. Coal is supplied by
20 trains each day. The efficiency of the station in converting the thermal energy released from
the coal to electrical energy is 26%.
A mass of 1.0 kg of coal will release 33 MJ of thermal energy when burnt.
Which mass of coal does each train bring?
A 2.5 × 104 kg B 6.3 × 104 kg C 1.5 × 106 kg D 3.0 × 107 kg

Respuesta :

skyluke89

The mass of coal needed is C) [tex]1.5\cdot 10^6 kg[/tex]

Explanation:

The average output of the power station is

[tex]P_{out}=3000 MW = 3\cdot 10^9 W[/tex]

Therefore we can find the average output energy [tex]E_{out}[/tex] per day using

[tex]P_{out}=\frac{E_{out}}{t}[/tex]

where

[tex]t=1d = 24\cdot 60\cdot 60=86,400 s[/tex] is the duration of one day

So we get

[tex]E_{out}=P_{out}t=(3\cdot 10^9)(86400)=2.59\cdot 10^{14} J[/tex]

The efficiency of the station in converting the energy is only 26%, so this means that the energy that must be produced in input by the coal per day is

[tex]E_{in}=\frac{E_{out}}{0.26}=1.0\cdot 10^{15}J[/tex]

This means that the input energy per train (20 trains) is

[tex]E=\frac{E_{in}}{20}=5.0\cdot 10^{13}J=50,000,000 MJ[/tex]

Since 1.0 kg of coal releases 33 MJ of thermal energy, we can write the following rule of three:

[tex]\frac{1.0 kg}{33 MJ}=\frac{m}{50,000,000 J}[/tex]

where m is the amount of coal needed for each train. Solving for m,

[tex]m=\frac{(50,000,000)(1.0)}{33}=1.5\cdot 10^6 kg[/tex]

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Lanuel

The amount (mass) of coal each train brings is equal to: C. [tex]1.5 \times 10^6 \;kg[/tex]

Given the following data:

  • Average power output = 3000 MW.
  • Number of trains = 20 trains
  • Efficiency = 26%.
  • Mass of coal = 1 kg
  • Energy released by a mass = 33 MJ

For a day:

Time = [tex]24 \times 60 \times 60 = 86400 \;seconds[/tex]

To determine the amount (mass) of coal each train brings:

First of all, we would solve for the average energy output of the power station by using the formula:

[tex]E_o = P_o \times time\\\\E_o = 3000 \times 10^9 \times 86400\\\\E_o = 2.59 \times 10^{14}\; Joules[/tex]

Next, we would determine the average energy input of the power station by using the formula:

[tex]E_{i} = \frac{E_o}{E_{ff}} \\\\E_{i} = \frac{2.59 \times 10^{14}}{0.26} \\\\E_{i} = 1.00 \times 10^{15}\; Joules[/tex]

For each train, the energy input is given by:

[tex]E_t = \frac{E_i}{number\;of\;trains} \\\\E_t = \frac{1.0 \times 10^{15}}{20} \\\\E_t = 5 \times 10^{13} = 50,000,000 \;MJ[/tex]

By direct proportion:

1 kg of coal = 33 MJ

X kg of coal = 50,000,000 MJ

Cross-multiplying, we have:

[tex]X = \frac{50000000 }{33} \\\\X = 1.5 \times 10^6 \;kg[/tex]

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