Science help!!

[laushik]

I need help with a word problem:

 

 A 750 W toaster and a 1200 W electric frying pan are plugged into the same 100 V outlet. How much will it cost to operate the two appliances at eight cents per kWh for one hour?

[dalekphalm]

1 minute ago, laushik said:

I need help with a word problem:

 

 A 750 W toaster and a 1200 W electric frying pan are plugged into the same 100 V outlet. How much will it cost to operate the two appliances at eight cents per kWh for one hour?

That's pretty easy to figure out.

 

First, the circuit (The outlet) doesn't care about individual draw. You can safely add the watts together: 1950W - fun fact, this would trip the breaker in your average North American outlet (They're rated at ~120V at 15A = 1800W total)

 

Now, you'll want to convert that into kW

 

1950 / 1000 = 1.95 kW

 

Then you just multiply that by your price: $0.08 kWh

 

1.95 kW * $0.08 kWh = $0.156

 

Or roughly 16 cents (rounded)

[laushik]

10 minutes ago, dalekphalm said:

That's pretty easy to figure out.

 

First, the circuit (The outlet) doesn't care about individual draw. You can safely add the watts together: 1950W - fun fact, this would trip the breaker in your average North American outlet (They're rated at ~120V at 15A = 1800W total)

 

Now, you'll want to convert that into kW

 

1950 / 1000 = 1.95 kW

 

Then you just multiply that by your price: $0.08 kWh

 

1.95 kW * $0.08 kWh = $0.156

 

Or roughly 16 cents (rounded)

I have another one:  an electric heater draws 1100 W of power. Electricity costs eight cents per kWh. How much does this cost to operate the heater three hours a day for 30 days?

[rmac52]

47 minutes ago, laushik said:

I have another one:  an electric heater draws 1100 W of power. Electricity costs eight cents per kWh. How much does this cost to operate the heater three hours a day for 30 days?

You should have no problem solving this using what dalekphalm provided.

[dalekphalm]

45 minutes ago, laushik said:

I have another one:  an electric heater draws 1100 W of power. Electricity costs eight cents per kWh. How much does this cost to operate the heater three hours a day for 30 days?

So rather then me just tell you the answer, you should try and work it out yourself and I'll check your work (Since knowing HOW you got the answer is much more useful).

 

So to find out how that works, we do a few things.

 

First, you've already got your wattage: 1100W.

 

kWh is a measurement of Power (Kilowatts or kW - 1000 watts = 1 kW) per hour. Kilowatts are measured in 1 hour increments, so a device that consumes 1kW will be measured at 1kWh - pretty simple!

 

So the first thing you'll want to do is convert your 1100W into kW. You just divide the number of W by 1000 to get kW.

 

Once you have your kW, you multiply it first by the number of hours per day (in your case 3), then you multiply this result by how many days you are measuring (In your case, 30).

 

Finally, you simply multiply that last number by your kWh charge (0.08). This will give you the total cost.

 

I will withhold the actual result for now.

 

Please reply with a post that breaks down each step, including the final result, and we'll confirm for you.

[AshleyAshes]

I feel like that this is a homework assignment where you've already been provided with the forumulas and the purpose of the questions are to determine if you understand how to plug numbers into the formulas to get results.  ...But instead you're asking the internet to understand it for you...

[dalekphalm]

3 minutes ago, AshleyAshes said:

I feel like that this is a homework assignment where you've already been provided with the forumulas and the purpose of the questions are to determine if you understand how to plug numbers into the formulas to get results.  ...But instead you're asking the internet to understand it for you...

I suspect the same thing - as this is SUPER BASIC high school math - like the kind of stuff you'd do in Grade 9 science (Or even Grade 7 Science maybe). I don't mind helping someone out with a difficult concept (The conversion from kW to kWh can be confusing to some), but I hope he learns how to do the calculations himself.

[laushik]

2 hours ago, dalekphalm said:

So rather then me just tell you the answer, you should try and work it out yourself and I'll check your work (Since knowing HOW you got the answer is much more useful).

 

So to find out how that works, we do a few things.

 

First, you've already got your wattage: 1100W.

 

kWh is a measurement of Power (Kilowatts or kW - 1000 watts = 1 kW) per hour. Kilowatts are measured in 1 hour increments, so a device that consumes 1kW will be measured at 1kWh - pretty simple!

 

So the first thing you'll want to do is convert your 1100W into kW. You just divide the number of W by 1000 to get kW.

 

Once you have your kW, you multiply it first by the number of hours per day (in your case 3), then you multiply this result by how many days you are measuring (In your case, 30).

 

Finally, you simply multiply that last number by your kWh charge (0.08). This will give you the total cost.

 

I will withhold the actual result for now.

 

Please reply with a post that breaks down each step, including the final result, and we'll confirm for you.

I got about $190... it seems wrong because it is kind of a lot for just 3 hrs everyday for 30 days. I really appreciate your help so far because the formula my teacher gave me is really wacky.

[dalekphalm]

1 minute ago, laushik said:

I got about $190... it seems wrong because it is kind of a lot for just 3 hrs everyday for 30 days. I really appreciate your help so far because the formula my teacher gave me is really wacky.

Can you write out your work, step by step? Because $190 is uh... super not correct. The number you're looking for is under $10 (Though $10 is not the correct answer).

[laushik]

What I did was divided 1100 W by 1000 to get 1.1 kw then I multiplied it by $0.08 and then I multiplied by 3 hours and got $0.264. I know that there is 720 hours in 30 days so I multiplied $0.264 by 720. That's what I did to get around $190.

[laushik]

1 hour ago, dalekphalm said:

Can you write out your work, step by step? Because $190 is uh... super not correct. The number you're looking for is under $10 (Though $10 is not the correct answer).

I know what I did wrong. I multiplied by 720 instead of 30, but shouldn't time always be calculated into hours?

[dalekphalm]

2 minutes ago, laushik said:

What I did was divided 1100 W by 1000 to get 1.1 kw then I multiplied it by $0.08 and then I multiplied by 3 hours and got $0.264. I know that there is 720 hours in 30 days so I multiplied $0.264 by 720. That's what I did to get around $190.

Your mistake was Multiplying by 720 - that's not correct. First off, you're not running the equipment for the entire 30 days (Or all 720 hours).

 

Since you're only operating the device 3 hours a day, you've already got your daily total (3). So from that point, you just need to multiply your daily total by how many days there are (30 - so 3 x 30 = 90).

 

So your machine is only running for 90 hours during that entire 30 day period. That's the figure you multiply with the cost per kWh.

 

Also, the other mistake you made was multiplying the cost per kWh with the daily figure, and not waiting until the very end. The number probably works out the same either way, but you can get rounding errors by doing things in the wrong order.

 

[dalekphalm]

2 minutes ago, laushik said:

I know what I did wrong. I multiplied by 720 instead of 30, but shouldn't time always be calculated into hours?

Yes of course time is calculated in hours. The problem is that you're calculating the wrong hours. By multiplying 3 x 30 (3 hours per day, for 30 days), you GET an hourly figure (90 hours).

 

Does that make sense?

[laushik]

1 minute ago, dalekphalm said:

Your mistake was Multiplying by 720 - that's not correct. First off, you're not running the equipment for the entire 30 days (Or all 720 hours).

 

Since you're only operating the device 3 hours a day, you've already got your daily total (3). So from that point, you just need to multiply your daily total by how many days there are (30 - so 3 x 30 = 90).

 

So your machine is only running for 90 hours during that entire 30 day period. That's the figure you multiply with the cost per kWh.

 

Also, the other mistake you made was multiplying the cost per kWh with the daily figure, and not waiting until the very end. The number probably works out the same either way, but you can get rounding errors by doing things in the wrong order.

 

Thanks so much!!! Yeah the order in which my teacher gave is very strange. 

[laushik]

4 minutes ago, dalekphalm said:

Yes of course time is calculated in hours. The problem is that you're calculating the wrong hours. By multiplying 3 x 30 (3 hours per day, for 30 days), you GET an hourly figure (90 hours).

 

Does that make sense?

Yes it does!

[dalekphalm]

3 minutes ago, laushik said:

Yes it does!

Good!

 

I get the impression your teacher didn't explain it very well.

 

BTW, can you tell me what the answer is now?

[laushik]

4 minutes ago, dalekphalm said:

Good!

 

I get the impression your teacher didn't explain it very well.

 

BTW, can you tell me what the answer is now?

I got $7.92. Also you stated how the order should be followed; this question requires only multiplication, why would order matter? I know you said rounding errors, but how?

[straight_stewie]

3 hours ago, dalekphalm said:

the kind of stuff you'd do in Grade 9 science (Or even Grade 7 Science maybe)

If you're in boy scouts you'll learn this around 6th grade while in WEBELOS. 

 

9 minutes ago, laushik said:

I got $7.92. Also you stated how the order should be followed; this question requires only multiplication, why would order matter? I know you said rounding errors, but how?

Order doesn't matter, after you convert Watts to kiloWatts. Order does matter because it's showing you what's actually going on in the problem, and allows you to relate maths to real life word problems.

Do you think $7.92 is correct? Why or why not?