What is the total energy in joules expended

  • Thread starter KillerZ
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  • #1
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Homework Statement



The power supplied by a certain battery is a constant 6 W over the first 5 min, zero for the following 2 min, a value increases linearly from zero to 10 W during the next 10 min, and a power that decreases linearly from 10 W to zero in the following 7 min. (a) What is the total energy in joules expended during this 24 min interval? (b) What is the average power in Btu/h during this time?

Homework Equations



[tex]P_{T} = P_{1} + P_{2} + P_{3} + P_{4}[/tex]

The Attempt at a Solution



I think I just find the power for the various times and add them up.

What I am confused with is when it says 6 W constant over 5 min does it mean:

[tex]P_{1} = (6 W)(5 min) = 30 W/min[/tex]

or does it mean:

[tex]P_{1} = 6 W[/tex]

then [tex]P_{2} = 0[/tex]

and I am lost on what to do about [tex]P_{3}[/tex] and [tex]P_{4}[/tex]
 

Answers and Replies

  • #2
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Remember,
[tex]P=\frac{W}{t}[/tex]

[tex]W=P\cdot t[/tex]

[tex]W_{tot}=W_{P_1}+W_{P_2}+W_{P_3}+W_{P_4}[/tex]

What you said about finding an equivalent power is simply false.
 
  • #3
CompuChip
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[tex]
P_{1} = 6 \text{ W}
[/tex]
is just "the power is constantly equal to 6W" expressed in a formula.

[tex]
P_{1} = 30 \text{ W/min}
[/tex]
is nonsense, unit-wise.

You seem a bit confused about power and energy. Can you explain (maybe first in words) what is power and energy and how they are related?
 
  • #4
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Ok. I think I have it:
[tex]W_{tot}=W_{P_1}+W_{P_2}+W_{P_3}+W_{P_4}[/tex]

[tex]W_{P_1} = (6 J/s)(300 s) = 1800 J[/tex]

[tex]W_{P_2} = (0 J/s)(120 s) = 0 J[/tex]

[tex]W_{P_3} = (55 J/s)(600 s) = 33000 J[/tex]

[tex]W_{P_4} = (55 J/s)(420 s) = 23100 J[/tex]

[tex]W_{tot} = 57900 J[/tex]
 
  • #5
671
2


Ok. I think I have it:
[tex]W_{tot}=W_{P_1}+W_{P_2}+W_{P_3}+W_{P_4}[/tex]

[tex]W_{P_1} = (6 J/s)(300 s) = 1800 J[/tex]

[tex]W_{P_2} = (0 J/s)(120 s) = 0 J[/tex]

[tex]W_{P_3} = (55 J/s)(600 s) = 33000 J[/tex]

[tex]W_{P_4} = (55 J/s)(420 s) = 23100 J[/tex]

[tex]W_{tot} = 57900 J[/tex]

Good, you're using the correct units now. Power is work per unit time.

You're wrong on the third and fourth parts of the process, though.

What does it mean that the power increases linearly from [tex]0 Watt[/tex] to [tex]10 Watt[/tex] over the course of 10 minutes?

Does it make sense that you have a power of [tex]55 Watt[/tex] in a process that at most produces [tex]10 Watt[/tex]?
 
  • #6
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Would it mean the slope say:

[tex]10/10 = 1 W/min [/tex]


[tex](1 W/min)(10 min) = 10 W[/tex]
 
  • #7
CompuChip
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Maybe this analogy helps: you are riding a bicycle and you increase your velocity from zero (stand-still) to 10 km/h in one minute. How far do you travel in one minute?
 
  • #8
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0.167 of a km?
 
  • #9
CompuChip
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Why?
(Numbers don't mean anything to me, tell me how you got it).
 
  • #10
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I said

[tex](10 km/h)(1 min / 60 min) = 0.167 km[/tex]
 
  • #11
CompuChip
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But you are starting from rest. Your answer would be correct if you were going 10 km/h the whole time.

It may help to draw a (v, t)-graph (you can also get the answer from that, have you learned how?)
Otherwise, can you tell me the average velocity over that first minute?
 
  • #12
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ok the area under a (v, t)-graph in the distance traveled. The (v, t)-graph would be a triangle so:

[tex](1/2)(b)(h) = (1/2)(1/60)(10) = 0.083 km[/tex]
 
  • #13
HallsofIvy
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Generally speaking, if something is increasing a at constant rate, its graph is a straight line and the "total" amount, the area under the line, can be gotten by finding the area of the trapezoid under the line. That is the same as saying that the average amount is just the average of the beginning and ending amounts. Here, the power " increases linearly from zero to 10 W during the next 10 min" so its starting value is 0 and ending value is 10. The average of those two numbers, and so the average power for the entire 10 minutes, is (0+ 10)/2= 5 and the "area under the graph", the area of a triangle, is 5(10)= 50.
 
  • #14
Delphi51
Homework Helper
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I see CompuChip is offline, so I'll try to fill in for now.
Your distance = area under the graph calculation is perfect. Note that you could also average the speed (0 + 10)/2 = 5 and do d = Vav*t = 5*1/60 = .083 km

You can use the same tricks to find the work - since W = P*t, when the power is not constant you can use Work = area under the Power vs time graph.
Or average the power and use W = Pav*t.
(the average method only works properly for a straight line graph)
 
  • #15
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so the last two would be:

[tex]W_{P_3} = (1/2)(10 J/s)(600 s) = 3000 J[/tex]

[tex]W_{P_4} = (1/2)(10 J/s)(420 s) = 2100 J[/tex]
 
  • #16
Delphi51
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Those look great!
Good luck converting to BTU. Wikipedia will probably have the conversion factor.
 
  • #17
CompuChip
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Thanks Delphi, nice teamwork there :)
 

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