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Low Power/Temp Heat Gun

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huckOhio
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I have a project using the MCP9808 temperature sensor.  Based on the readings from the sensor different electrical devices will turn on or off (fans, heaters, etc.).  The temperature ranges I am testing against is 70-80 degrees for summer and 40-30 degrees for winter (for winter I use a canned air duster and turn the can up-side-down).  I am testing the project in my basement which is at 68 degrees.  Currently, to simulate warmer temperatures (70-80 degrees) I am rubbing my finger against my jeans to get it hot and then I touch the temp sensor.  Now that I have worn away my finger print, I was wondering if  anyone might have a better idea (besides turning up the furnace).  I googled low power heat guns and the lowest still puts out between 200-300 degrees.  Not sure if such a piece of test equipment exists...but I thought I would ask the forum.  Thanks!


   
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Will
 Will
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Posted by: @huckohio

I am rubbing my finger against my jeans to get it hot and then I touch the temp sensor.  Now that I have worn away my finger print, I was wondering if  anyone might have a better idea (besides turning up the furnace).

How about a nice hot cup of tea 🙂

 

Experience is what you get when you don't get what you want.


   
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Ron
 Ron
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@will @huckohio A HOT cup of tea is several times hotter than 75. Just fill a cup with water from the tap that is on the cool side of tepid (59-98), use a candy thermometer to get it to 75ish.

"Don't tell people how to do things. Tell them what to do and let them surprise you with their results.” - G.S. Patton, Gen. USA
"Never wrestle with a pig....the pig loves it and you end up covered in mud..." anon


   
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Ron
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Posted by: @huckohio

I googled low power heat guns and the lowest still puts out between 200-300 degrees.  Not sure if such a piece of test equipment exists...but I thought I would ask the forum.  Thanks!

A hair dryer on low held far enough away and a thermometer to measure the air around the sensor.

 

"Don't tell people how to do things. Tell them what to do and let them surprise you with their results.” - G.S. Patton, Gen. USA
"Never wrestle with a pig....the pig loves it and you end up covered in mud..." anon


   
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Will
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Posted by: @zander

FYI @huckohio A HOT cup of tea is several times hotter than 75. Just fill a cup with water from the tap that is on the cool side of tepid (59-98), use a candy thermometer to get it to 75ish.

But then you don't get to drink the tea ...

 

Experience is what you get when you don't get what you want.


   
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Will
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Posted by: @zander

Posted by: @huckohio

I googled low power heat guns and the lowest still puts out between 200-300 degrees.  Not sure if such a piece of test equipment exists...but I thought I would ask the forum.  Thanks!

A hair dryer on low held far enough away and a thermometer to measure the air around the sensor.

Well, maybe YOU and @HhuckOhio have enough hair left to need a hair dryer but I certainly don't (have one) 🙂

 

Experience is what you get when you don't get what you want.


   
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Ron
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@will He could do both, make a tea and get some tepid water one does not preclude the other.

"Don't tell people how to do things. Tell them what to do and let them surprise you with their results.” - G.S. Patton, Gen. USA
"Never wrestle with a pig....the pig loves it and you end up covered in mud..." anon


   
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Ron
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@will You seen my scalp, I also don't have one but my wife does. If he is a bachelor like you he is SOL unless he has a heat gun or heat shrink hot air gun. Again, low and far enough away to yield the desired result.

"Don't tell people how to do things. Tell them what to do and let them surprise you with their results.” - G.S. Patton, Gen. USA
"Never wrestle with a pig....the pig loves it and you end up covered in mud..." anon


   
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Will
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Posted by: @zander

@will He could do both, make a tea and get some tepid water one does not preclude the other.

I disagree, hot tea has utility in that it is drinkable and, since it will eventually become tepid, will eventually replace the tepid water anyway.

 

Experience is what you get when you don't get what you want.


   
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Ron
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@will True, it just depends if he wants to wait for the tea to cool or not.

"Don't tell people how to do things. Tell them what to do and let them surprise you with their results.” - G.S. Patton, Gen. USA
"Never wrestle with a pig....the pig loves it and you end up covered in mud..." anon


   
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Will
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Posted by: @zander

If he is a bachelor like you he is SOL unless he has a heat gun or heat shrink hot air gun.

Or a kettle and tea bag 🙂

I'm going to stop now in case somebody has a reasonable answer for @HuckOhio's question 🙂

 

Experience is what you get when you don't get what you want.


   
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huckOhio
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LOL.....all are excellent suggestions 😀 .  @Zander Unfortunately, my hair and I have parted ways some time ago.  But, I do believe we have an old hair dryer laying around somewhere.  

Thanks all.  Enjoyed the comments.  


   
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frogandtoad
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@huckohio

Have you considered driving some current through a power resistor via a pot to regulate it?

You might also like to investigate the use of a Peltier Device (Bill has made a video of this in the past), to generate the heat and cooling?  I'm pretty sure that you can regulate the heat via a pot with both of these approaches, though I don't know the mathematical formula off hand!  Perhaps someone like @davee can provide the formula to convert voltage/watts to heat?

Otherwise, you could just point a cheap infrared temperature gun at either, and adjust your pot to desired temperature?

Finally, pumping some current through a small "Nichrome Wire" in the same way, and building your own thermal element could be a solution?

Just some food for thought.

Cheers


   
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DaveE
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Hi @frogandtoad,

Re: Perhaps someone like @davee can provide the formula to convert voltage/watts to heat?

Sadly, no such 'general formula' exists.

If you apply a source of heat to an object, then in 'real' circumstances, some of the heat energy will be absorbed by the object, causing its temperature to increase, and some of the heat energy will be dissipated into the environment.

For a specific amount particular material, it is possible to look up the number of Joules needed to raise its temperature by 1 °Celsius ... and hence use that proportionality for a particular situation ...

   e.g. 1 g of liquid water at about 20°Celsius requires 4.186 Joules to heat it by 1 °Celsius

        Noting that a 1 Watt power flow is equal to 1 Joule per second...

The value of 4.186 J/g°C is only an 'average' measured figure for water at a 'standard temperature' ... another material will have a different value.

And, this assumes no heat is lost to the surroundings ... which is an impossibility, although insulation, etc. may minimise the heat loss. Clearly it depends on the physical situation which is usually complex to determine and it is not a simple formula.

Furthermore, if the water changes state as a result of the heating e.g. ice <-> liquid or liquid <-> vapour, then a large amount of energy is involved in those transitions.

---------

A quick review of this thread leaves me a bit confused as to what @huckohio is trying to achieve. If it is calibration of a sensor, then a common approach is to have a 'bath' of liquid whose temperature can be altered, and a 'calibrated' thermometer. Then compare the thermometer reading with the sensor reading over a range of temperatures.

For temperatures in the range 0°C to 100°C, this might be simply a small bowl/saucepan of water, whose temperature could be varied by mixing water from the tap, boiling water from the kettle, and ice cubes as appropriate to produce a range of temperature points. Of course, water conducts electricity which can affect the reading, so care is needed to keep the connections to sensor, and possibly the sensor itself, dry.

Hope this helps, Dave


   
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Inst-Tech
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@huckohio, @davee

 

Posted by: @davee

Hi @frogandtoad,

Re: Perhaps someone like @davee can provide the formula to convert voltage/watts to heat?

Sadly, no such 'general formula' exists.

If you apply a source of heat to an object, then in 'real' circumstances, some of the heat energy will be absorbed by the object, causing its temperature to increase, and some of the heat energy will be dissipated into the environment.

For a specific amount particular material, it is possible to look up the number of Joules needed to raise its temperature by 1 °Celsius ... and hence use that proportionality for a particular situation ...

   e.g. 1 g of liquid water at about 20°Celsius requires 4.186 Joules to heat it by 1 °Celsius

        Noting that a 1 Watt power flow is equal to 1 Joule per second...

The value of 4.186 J/g°C is only an 'average' measured figure for water at a 'standard temperature' ... another material will have a different value.

And, this assumes no heat is lost to the surroundings ... which is an impossibility, although insulation, etc. may minimise the heat loss. Clearly it depends on the physical situation which is usually complex to determine and it is not a simple formula.

Furthermore, if the water changes state as a result of the heating e.g. ice <-> liquid or liquid <-> vapour, then a large amount of energy is involved in those transitions.

---------

A quick review of this thread leaves me a bit confused as to what @huckohio is trying to achieve. If it is calibration of a sensor, then a common approach is to have a 'bath' of liquid whose temperature can be altered, and a 'calibrated' thermometer. Then compare the thermometer reading with the sensor reading over a range of temperatures.

For temperatures in the range 0°C to 100°C, this might be simply a small bowl/saucepan of water, whose temperature could be varied by mixing water from the tap, boiling water from the kettle, and ice cubes as appropriate to produce a range of temperature points. Of course, water conducts electricity which can affect the reading, so care is needed to keep the connections to sensor, and possibly the sensor itself, dry.

Hope this helps, Dave

I found this Watts to heat calculator https://www.omnicalculator.com/physics/watts-to-heat .. this may not be what @huckohio is looking for, but it's interesting!...lol

A look at the peltier effect is also a option as @frogandtoad eluded too.  portable heat/cool ovens are also available as testing equipment, but are very expensive and probably not an option..I personally like the hair dryer solution as it appears to be the most practicable. 

good luck with your project @huckohio..

regards,

LouisR

 

LouisR


   
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