# Measuring watts

1. Oct 30, 2005

### furtivefelon hi, i have another question i can't get, the problem is:

We want to boil one liter of water in a light thin-walled 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. 3. Oct 30, 2005

### Chi Meson OK. at "500 W-W", 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).

4. Oct 30, 2005

### Pengwuino Can you explain what this means? Do you mean how many watts are being generated from the heater?

5. Oct 30, 2005

### furtivefelon that's the exact wording of the question given to me.. However, from my understanding, that the watt inscribed on the cover of the heater isn't the "real" watt, rather, a factory specific watt, so i'm guessing they want me to find the conversion between this "wonderland" watt and normal watt..

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

6. Oct 30, 2005

### Chi Meson Hey, was my last post not enough of a hint? You ought to be done by now 7. Oct 30, 2005

### furtivefelon erm.. since we only started learning past Newton machanics, i have to do my own research, just started reading on Thermodynamics (should've done that a long time ago)..

Just wondering, what does IOW mean in ur first post?

and thanks alot for responding i really appreciate it 8. Oct 30, 2005

### furtivefelon 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. Oct 30, 2005

### Pengwuino Where is work being done?

10. Oct 30, 2005

### Chi Meson 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?

11. Oct 30, 2005

### furtivefelon 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)

12. Oct 30, 2005

### furtivefelon Now looking at the physics book i've got, i see that they give two equations for calculating heat, one is cm(delta T), the second one is c(delta T).. I'm not sure about the practical differences between "specific heat capacity" and "heat capacity"..

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 13. Oct 30, 2005

### Chi Meson power is also Q/delta t. I already told you that.

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.

14. Oct 30, 2005

### furtivefelon sorry about that, i wrote it in response to Pengwuino, and after reading your post, i understood everything :D I didn't make the connection that both are energy transfers, now i did, both quantity makes sense now :D

15. Oct 30, 2005

### Pengwuino Yup and...

$$\Delta E = \Delta Q + \Delta W$$

Or at least it better be....

16. Oct 30, 2005

### Chi Meson It's not just a good idea, you know. It's the Law.

17. Oct 30, 2005

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

18. Oct 30, 2005

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

19. Oct 30, 2005

### Pengwuino Oh it won't probably. It's just to point out that heat and work are both forms of energy.

20. Oct 30, 2005

### furtivefelon So to calculate joules per second, i can use Q=cm(delta T)/delta t right? I know from density of water that 1 liter = 1kg of water.. and 20 minutes = 1200 seconds, therefore, Q=(4.186 J/g*C)(1000g)(2)/1200s = 6.97Watt..

Though, if i'm right so far, i don't really know the delta T is when the heater heated the water up, as i don't know from the quesion the initial temperature the water started out with.. is there any other way to approach this?

21. Oct 30, 2005

### Chi Meson You don't need to. At 60 degrees the heat was leaving the container at the same rate (nearly) as it did for the next 20 minutes as it cooled off. That means this is the rate that heat was going in (heat in = heat out means no change in T). So you have a way of finding the actual wattage because this equals the rate of heat leaving at that time.