Simple units conversion question

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Homework Help Overview

The discussion revolves around unit conversions, specifically converting energy from Joules to kilowatt-hours (kWh). Participants are examining the appropriate conversion factors and the reasoning behind whether to multiply or divide by certain values during the conversion process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the confusion surrounding the conversion process, particularly the use of 3600 in the calculations. There are attempts to clarify the reasoning behind multiplying or dividing by conversion factors, with examples provided to illustrate unit cancellation and the importance of maintaining consistent units.

Discussion Status

The conversation is ongoing, with participants sharing insights and examples to clarify the conversion process. Some guidance has been offered regarding the expectations of the numerical outcomes when converting between units, but no consensus has been reached on the best approach yet.

Contextual Notes

There is a note regarding the appropriateness of thread titles in the forum, indicating a preference for clarity in categorizing the level of the question being asked.

bonbon22
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Homework Statement
Show that the cost of this wasted energy will be about £4, if electrical energy is
charged at 20 p per kWh. given that total energy wasted energy is 7.5(6) × 107 (J)
Relevant Equations
no equations
as it wants the units kWh does that not mean you times by 3600 to go from s --> h and to go from W to kilowatt --> you divide the value by 1000

yet they divided by 3600 to get answer
 
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Unit conversions can be confusing, and it's easy to mix up whether you should multiply or divide by the conversion factor. The best way is to carry along the units in your calculation, canceling them along the way , like this:
[tex]1\, kWh = 1000 \, W h = 1000 \frac{J}{s}h \times 3600 \frac{s}{h} = 3.6 \times 10^6 J[/tex]
so:
[tex]7.5 \times 10^7 J = \frac{7.5 \times 10^7 J}{3.6 \times 10^6 \frac{J}{kWh} }= 20.8\, kWh[/tex]
 
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bonbon22 said:
yet they divided by 3600 to get answer
As shown by @phyzguy it is best to carry units along in your calculations. Units on the LHS and RHS of each equation should match. A helpful trick for doing units conversions is to "multiply by one", where the "one" has the units you want to end up with in the numerator, and the units you want to cancel out in the denominator.

Like this: Convert 60 km/hr into m/s:

[tex]1 = 3600 \frac{s}{hr} = 60 \frac{s}{min} * 60 \frac{min}{hr}[/tex]

[tex]60 \frac{km}{hr} * \frac{1 hr}{3600 sec} * \frac{1000 m}{1 km} = 16.7 m/s[/tex]
 
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It also helps to have an idea of whether you should get a bigger or smaller number after the conversion. For example, 1 km and 1000 m both describe the same length, but because it takes a bunch of meters to make a kilometer, i.e., a kilometer is longer than a meter, you need more meters (1000) than kilometers (1) to specify the same length.

In your case, you want to convert from Joules to kWh. A joule is a watt-second, which is smaller than a watt-hour, so for a given amount of energy, there would be more watt-seconds than watt-hours. Therefore, going from joules to watt-hours, you should expect to get a smaller number. Given the choice of multiplying by 3600 or dividing by 3600, which calculation is going to yield a smaller number? Dividing. So you should divide by 3600, not multiply.
 
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@bonbon22, please don't add "A level" to your thread titles, even if it might be technically accurate, because it can be confusing to readers here. At this forum, titles can be tagged as B-, I-, and A-level, indicating that the question is at one of these levels: Basic (high school), Intermediate (undergraduate), or Advanced (graduate).
 
Mark44 said:
please don't add "A level" to your thread titles
LOL, I didn't even catch that! :smile:
 

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