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Measuring watts.. 
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#1
Oct3005, 10:29 AM

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hi, i have another question i can't get, the problem is:
We want to boil one liter of water in a light thinwalled vessel. For this purpose, we start to heat water by an electric heater, with the label "500W. Made in Wonderland". However, after the temperature becomes 60*C, it stops increasing. We become bored wtih waiting and switch off the heater. Our measurement show that during the first 20 minutes water becomes 2 degrees cooler. How many watts contains the "Wonderland's watt"? I've read a long time on charges, electric field, current and resistance.. I still can't get how the approach the question.. can someone give me a hint as how to get started? Thanks alot :D 


#2
Oct3005, 03:41 PM

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OK. at "500 WW", equilibrium is reached at 60 degrees. IOW, the heat into the water equals the heat out of the water (through the thin walls of the container).
Through these thin walls, heat is leaving at a rate such that after 20 minutes, cm(deltaT) joules of heat have left (delta T is 2 degrees). 


#3
Oct3005, 03:49 PM

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#4
Oct3005, 04:14 PM

P: 30

Measuring watts..
Though, donno if my intepretation is right.. If it's wrong, please point it out :D 


#5
Oct3005, 04:18 PM

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Hey, was my last post not enough of a hint? You ought to be done by now



#6
Oct3005, 04:23 PM

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Just wondering, what does IOW mean in ur first post? and thanks alot for responding i really appreciate it 


#7
Oct3005, 04:49 PM

P: 30

there is one thing i don't understand while in my reading.. The unit of Work is Joules, and the unit of Heat is joules as well.. If those two are interchangeable, then this problem becomes quite easy.. what exactly is the difference(if any) exists between work and heat?



#9
Oct3005, 05:08 PM

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IOW = "in other words"
work is the transfer of energy through mechanical means (by way of forces exerted through distances). heat is the transfer of thermal energy. Since both are energy transfers, both are measured in joules. Power (rated in watts) is the rate of transfer of energy. This can be heat/time or work/time. Either way, a watt is a joule per second. You thin walled heating thingy has heat going in at a certain number of joules per second. At 60 degrees, it doesn't increase in temperature anymore because heat is leaving at the same rate. So joules per second in = joules per second out. The second part of the question allows you to calculate the number of joules out in 20 minutes. How many joules out per second then? 


#10
Oct3005, 05:09 PM

P: 30

mmm.. i got that notion after looking at the power equation, P = W/delta t, as there isn't exactly any relationship that i can find between power and heat.. Further, as work and heat has the same unit, and you can change temperature in a body by doing work to it (such as compressing air), also heating the air by bunsen burner would produce indistinguishable result (if compressing, in the beginning, they have different volume, if heat, in teh beginning, they have the same volume..)
so i thought i'd ask is there any place that work and heat is different? (if you need any clearification on my reasoning in the previous paragraph, please alert me :D) 


#11
Oct3005, 05:18 PM

P: 30

The only mass i know from the question is the mass of the water (1 L = 1 Kg by density).... I'm not sure if that's the right mass to use in cm(delta T) sorry if i sound very confused right now, it's just that i'm learning alot of stuff right now and i'm trying to sort everything out bit by bit.. Thanks alot for your patience 


#12
Oct3005, 05:19 PM

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the entirely of your last post requires clarification. "any place that work and heat is different?" But they are different things. They are the same only when you look at individual molecular collisions. 


#13
Oct3005, 05:25 PM

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#14
Oct3005, 05:45 PM

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Yup and...
[tex]\Delta E = \Delta Q + \Delta W[/tex] Or at least it better be.... 


#15
Oct3005, 06:18 PM

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#16
Oct3005, 06:20 PM

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Im very skeptical everytime i put out an equation since im never really 100% sure what im saying is right lol



#17
Oct3005, 07:10 PM

P: 30

mmm.. i don't quite understand how that equation will help me, that is, finding the difference in energy..



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