If I use a vehicle battery connected to jumper cables, which is then connected to the heating element of a kettle, which is then placed in play sand, in a metal container, will the sand then heat up?
yes.
assuming it is a normal uk kettle it is normally 2 to 3kw at 230v
assuming it is 3kw
3000 / 230 x 12 =150watts
so it would warm the sand but not very quickly as it will only give out 1/20th of its normal power output as it is being run at 1/20th of its normal voltage.
assuming it is a normal uk kettle it is normally 2 to 3kw at 230v
assuming it is 3kw
3000 / 230 x 12 =150watts
so it would warm the sand but not very quickly as it will only give out 1/20th of its normal power output as it is being run at 1/20th of its normal voltage.
I was thinking 150W is still a lump of power
Using online calculator Watts to Heat Calculator - https://www.omnicalculator.com/physics/watts-to-heat
To heat 1Kg of sand by 10 DegC with 150W will take just under 1 minute (still sounds unlikely - mistake?)
I suspect the thermal conductivity of sand is not great, so it might ultimately get quite hot.
Whereas 1Kg (1L) of water with the same 150W would take nearly 5 minutes to rise 10 DegC!
I hope the mention of 'play sand' doesn't suggest it might be played with, as I suspect this could ultimately get dangerously hot in the vicinity of the element?
I don't think it is a mistake, would not be surprised if sand has a specific heat capacity of 0.2 x that of water.
I don’t think the cup of tea is going to taste very nice….. and maybe a bit cold
Engineer1969
Deleted account
Just a word of warning...
Be careful and mindful of not shorting out the battery if the jumper cables or battery are inside of a metal container
Make sure the contacts are completely isolated from the metal container even in the event of someone accidentally moving / kicking it or it could cause the battery to catch fire
That calculation is incorrect
3000 / 230 is the formula for amps (I=P/V) which you are then multiplying by 12 to equal 150 amps not 150 watts
If the load was 3000 watts at 12v then the correct formula for amps would be 3000w / 12v = 250a
However it would not be 3000w at 12v. This should be calculated based on the resistance of the element not the wattage which will change with the voltage
So also assuming 3000 watts at 230v
R = V² / P
52900v / 3000w = 17.6Ω
So at 12v
P = V² / R
144v / 17.6Ω = 8.2w (or 0.68a)
I think it would take a long time to heat up at 8.2 watts
I = P / V ......... 3000w at 230v =13.04A
R = V / I ........ 230 / 13.04 = 17.63 Ohm
I = V / R ........ 12 / 17.63 = 0.68A
P = V * A ........ 12 * 0.68 = 8.16W
As you said before, My original calculation was wrong.
I have expanded it so others can understand (particularly those new to the field)
@Engineer1969 Well done and thank you for pointing out the mistake.
its not often I make a error as fundamental as that.
R = V / I ........ 230 / 13.04 = 17.63 Ohm
I = V / R ........ 12 / 17.63 = 0.68A
P = V * A ........ 12 * 0.68 = 8.16W
As you said before, My original calculation was wrong.
I have expanded it so others can understand (particularly those new to the field)
@Engineer1969 Well done and thank you for pointing out the mistake.
its not often I make a error as fundamental as that.
We've had similar mistakes on here referring to electric showers.
Shower is rated at 10.5kW at 240V = 43.75A , therefore at 230V it would draw 10500/230A = 45.65A, whereas in reality, the current is less at 230V.
Shower is rated at 10.5kW at 240V = 43.75A , therefore at 230V it would draw 10500/230A = 45.65A, whereas in reality, the current is less at 230V.
Hello mate,
Thankyou for your advice and no! For a bit of context I’m in the army and have been tasked to create something, that is portable, quick to set up and can be hot enough to be seen on a thermal sight.
Thankyou for your advice and no! For a bit of context I’m in the army and have been tasked to create something, that is portable, quick to set up and can be hot enough to be seen on a thermal sight.
You could use a battery and a lamp, wrap the lamp in foil if you don't want the light to be seen.
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Pop into the vehicle workshop and blag a headlight bulb - they'll regularly be replacing them because dipped beam has blown, while main beam is fine. Or just blag a new one.
While you are there, blag some cable, Croc clips, and a fuse.
Then pop down to the kitchen and ask for a largish tin wih a lid - post scran removal. Make a hole, fix the bulb into the hole with light inside the tin.
You now have a 55W (maybe more) heater inside a tin. The tin will get warm if uncovered, hot if covered.
Years ago, dad made something like that as a bed warmer, using a 10lb trade tin of Horlicks - I think it only needed something like a 15W bulb.
While you are there, blag some cable, Croc clips, and a fuse.
Then pop down to the kitchen and ask for a largish tin wih a lid - post scran removal. Make a hole, fix the bulb into the hole with light inside the tin.
You now have a 55W (maybe more) heater inside a tin. The tin will get warm if uncovered, hot if covered.
Years ago, dad made something like that as a bed warmer, using a 10lb trade tin of Horlicks - I think it only needed something like a 15W bulb.
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