01 May 04
1 edit
Originally posted by flexmoreThe numbers are from memory, so correct me if I'm wrong.
if a 25 watt light globe is used for 1 hour, then how much hot water could have been produced instead?
(assume the water goes from 10C to 75C)
It takes 4200J to heat up 1kg of water by 1C, so it would take 273 000 J to heat 1kg up by 65C. A 25W light bul running for 1 hour uses 25x3600 = 90 000 J. So you could heat up 90/273 l of water, which is roughly the contents of a can of coke or a mug of coffee.
Here's a similar question:
A faulty kettle is filled with freezing cold water and left on. It heats up and boils dry over the course of an hour. Ignoring the heat capacity of the kettle, roughly what is the mean temperature of the water in the kettle over this hour?
07 May 04
Originally posted by Acolyteseveral factors comeinto play.
A faulty kettle is filled with freezing cold water and left on. It heats up and boils dry over the course of an hour. Ignoring the heat capacity of the kettle, roughly what is the mean temperature of the water in the kettle over this hour?
first, is there any ice in the water? that will slow down the heating process until the ice is gone.
also, is the kettle covered? it will heat faster if it is.
third, what is the heating capability of the stove?
i'll assume that the kettle contains four litres and a 5kw heating element. if you heat the water from 0 to 100c, it takes 16800 j per degree (if your figures are right, and they usually are), but of course it takes the same amount (i think) to melt any ice. we'll do it two ways, one with no ice and the other with 500g ice.
for the former, you would need 1.68 Mj and the heating element puts out 18 Mj/hr so it will heat the water in barely under a tenth of an hour or just over five minutes. since the temperature will remain at 100c until the water is gone, and it has been stated that this does not happen until the hour is over, the temperature would reach and maintain 100c for just over 54 minutes. since it is 100 for those times and an average of 50 the rest (total of 5700 degree-minutes, over a 60-minute interval), the mean would be 95.
for the second part, i would think the time difference would be fairly small since the mass of the ice is an eighth of the total, but will add about a minute. you will have one minute of 0, the same rapid rise to 100, and that sustained for 53 minutes and the mean will be 93+ (5600 deg-min).
i don't know the heat of conversion though, so i may be way off on that.
08 May 04
Originally posted by BarefootChessPlayerNo ice, and I'm assuming the kettle is perfectly insulated; the power and capacity of the kettle are irrelevant, as you already have enough information (you can look up the latent heat of vaporisation of water on the internet).
several factors comeinto play.
first, is there any ice in the water? that will slow down the heating process until the ice is gone.
also, is the kettle covered? it will heat faster if it is.
third, what is the heating capability of the stove?
i'll assume that the kettle contains four litres and a 5kw heating element. if you heat the water fr ...[text shortened]... + (5600 deg-min).
i don't know the heat of conversion though, so i may be way off on that.[/b]
- Joined
- 26 Apr 03
- Moves
- 26771
09 May 04
1 edit
Originally posted by AcolyteA fine question, let's give it a go
[bA faulty kettle is filled with freezing cold water and left on. It heats up and boils dry over the course of an hour. Ignoring the heat capacity of the kettle, roughly what is the mean temperature of the water in the kettle over this hour?
...[text shortened]... look up the latent heat of vaporisation of water on the internet).[/b]
The latent heat of vaporisation is 2260 Kj/Kg [1]
We shall divide the experiment into the heating from 0 to 100 phase, which takes H seconds, and the evaporation phase, which takes E seconds. We also know that H + E = 3600, and so H = 3600 - E
The average temperature during the heating phase is 50c, and the average temperature during evaporation is 100c, therefore the average temperature over the whole experiment is:
(50H + 100E)/(H+E), from which we can eliminate H and simplify to give:
average temperature = 50+e/72 [2]
If we have to evaporate X kilogrammes of water then from [1] we know our kettle outputs 2260X Kj per E seconds
However the kettle must also heat up the water to 100 in H,
the specific heat capacity of water is 4.184Kj per degree c per Kg
therefore the kettle must output 418.4X Kj per H seconds
we can therefore form the equation:
418.4X/(3600-E) = 2260X/E
cancelling X from both sides and rearranging:
E = 3037.6 (approx)
and so from [2] the average temperature is 92.33 degrees c
16 May 04
I am not sure that "average" temperature is in question, rather, the "mean" temperature. . . which is which temperature occours more times. 🙄
😕 Consider the following: 😕
1) I feel that if you start with IceWater, the temp. is 1ºC
2) By applying heat, the temp. steadily goes up, not staying at any one º for too long. until....
3) Once the (now) water reaches 100ºC, it boils. (more than 100º does not really matter since we are not measuring steam temp right?)
4) The heat must now stay at least 100ºC for a long time in order for the water to boil away.
THEREFORE, the "mean" temperature is 100ºC since it stays at this temp. the longest. 😏
Anyhow, there are other variables...
1. How much freezing water do we start with?
2. What elevation are we at? <-- big question since it matters