Thermodynamics. How to precisely measure energy released by body.

[Sensei]

Thermodynamics. How to precisely measure energy released by body.

Imagine circuit with resistor (heating element) with resistance R.
There is also voltmeter, ampere meter connected.
[math]I=\frac{U}{R}[/math]
and
[math]P=I*U=I^2*R=\frac{U^2}{R}[/math]
R is constant (would be hard to control it in heating element underwater).
Heating element is placed in water (or other liquid) with mass m, and volume V.
There is also precise thermometer, stopwatch, and magnetic stirrer to uniformly spread energy in liquid.

At the beginning temperature is ambient T₀ read from thermometer. Say 25 C.
(ambient temperature is when amount of energy released by body equals amount of energy received from environment)

We're heating it to T₁, say 50 C.

Electrons are accelerated, and while passing through heating element, they lose their kinetic energy and resistor is heated, and indirectly liquid.
We read voltage drop on heating element from voltmeter. And read quantity of electrons Q/e, Q=I*t from ampere meter and stopwatch.

E_input=I*t*U

We can make equation to calculate energy needed to heat liquid with mass m, from T₀ to T₁:
[math]4.1855 \frac{J}{K*g} * m * (T_1-T_0) = ~ I*t*U[/math]

Calculate amount of energy needed to heat 1 gram of substance by 1 K:
[math]4.1855 \frac{J}{K*g} = ~ \frac{I*t*U}{m * (T_1-T_0)}[/math]

Calculate amount of energy needed to heat 1 mole of substance (water in example):
[math]4.1855 \frac{J}{K*g}*18.015\frac{g}{mol} = ~ \frac{I*t*U}{\frac{m}{18.015\frac{g}{mol}} * (T_1-T_0)}[/math]
[math]75.4 \frac{J}{K*mol} = ~ \frac{I*t*U}{n * (T_1-T_0)}[/math]
(etc. etc.)

It's idealized equation: when no energy is lost by emitting photons (in this case in microwave and infrared region of spectrum), nor by air gas particles taking part of energy.

I was thinking how to precisely measure amount of energy released by body to environment and influencing our idealized equation.

And came up with idea to use voltage/current regulator.

If we supply more energy to our system, than is released, temperature will increase.
T_current must by higher than ambient temperature T₀.
E_input>E_radiated, T_new>T_current, T_current>T₀

If we supply less energy to our system, temperature will decrease (only to T₀).
E_input<E_radiated, T_new<T_current, T_new>=T₀

But if we use voltage/current regulator to precisely set value, with precision counted in mV, or uV, or better, temperature will be constant. Neither grow, nor fail.
E_input=E_radiated, T_new=T_current=const

And we will know exactly how much is radiated as photons or heat.

This experiment could be performed at couple different temperatures, increased by f.e. 10 C.
And receive set of data f.e.:
E_input(T₀+10)
E_input(T₀+20)
E_input(T₀+30)
....
E_input(T₀+70)

Which will be placed on graph with temperature in x axis, and Energy/Current/Voltage/Power in y axis.

Second experiment: change area which has contact with air, while mass and volume remain the same.
To compare how results changes with area exhibited to air.

Third experiment: compare plain liquid graph, with graphs obtained with liquids which have metal radiator different shape and size.
Thus we will be able to measure efficiency of different radiator models.

Fourth experiment: repeat everything, with different liquid than water. And compare data with water on graph.

Best Regards!

[swansont]

What you've described is a kind of calorimeter.

https://en.wikipedia.org/wiki/Calorimeter

[Sensei]

What you've described is a kind of calorimeter.

https://en.wikipedia.org/wiki/Calorimeter

 

Sort of.

 

Quote from link:

"It does not account for the heat loss through the container or the heat capacity of the thermometer and container itself."

Mine device is designed for measuring this loss.

 

Typical calorimeter is insulated as best as they can make it. To keep heat inside of device.

While in mine device insulation is not needed. Because what is radiated away is what we want to measure. Not energy released during burning f.e. (as there is nothing to burn in it).

[swansont]

Which means you are proposing an improvement, by calibration. It's still a calorimeter.

[John Cuthber]

I rather suspect that I could buy "off the peg" a calorimeter that's accurate enough to measure the heat added to your system by the magnetic stirrer, but that you have not accounted for.

That's the sort of system that lets you measure the 4.1855 KJ/Kg K which you need to use in your system.

[Sensei]

I rather suspect that I could buy "off the peg" a calorimeter that's accurate enough to measure the heat added to your system by the magnetic stirrer,

I really would like to see it.

It might be accurate enough to measure +-0.050 J difference in burned sample, but it won't measure it by itself.

If you turn off stirring, temperature won't be uniform anymore, but create gradient of temperatures inside of device (sealed) water container. Making reading from thermometer not reliable anymore.

Even my hot tea that I just measured has 57.5 C at near top, and 54.5 C at bottom (measured by my -50 C...+300 C thermometer with +-0.1 C precision)

 

If stirrer engine has U=5 V voltage drop, and I=0.05 A, that's 0.05 J/s introduced to system at worst maximum.

(just example data for electric engine, the real data would be from 2nd voltage/ampere meters)

 

If volume of water in container is 100 mL = 100 g, 4.1855 J/K*g * 100g * 1 K = 418.55 / 0.05 = 8371 seconds wait for increase water temperature by 1 C..

 

but that you have not accounted for.

But is it really needed? You should rethink it.

If stirrer is turned on, its engine has pretty much the same power when stirred liquid has temperature T₀=25 C, and when T₁=50 C, or when it has T₂=75 C.

It's always introducing exactly (+-) the same amount of energy to the system, regardless of whether heating element is turned on or off.

 

That's the sort of system that lets you measure the 4.1855 KJ/Kg K which you need to use in your system.

 

You also misunderstood what I am interested in measuring in my device, in the first place.

4.1855 J/K*g is just example. I don't need to measure it. But can measure it, if I would like to.

But what I wanted to really measure is amount of energy radiated away by body.

 

Again.

We have T₀=25 C initial temperature.

Device is turned on. Voltage on voltage regulator slider is adjusted to reach temperature f.e. 50 C.

If T is still going up, voltage is adjusted to smaller value, if T is going down too fast, it's adjusted in reverse direction.

Adjust until we reach equilibrium T=const, and some U on regulator.

When it does not change, we can read amount of energy radiated away by body.

 

Repeat the same with T=70 C, or any other.

The key is to have constant temperature reading.

Amount of energy introduced to system by stirrer is constant, regardless at which T we're currently working.

Amount of energy from environment introduced to system is constant, regardless at which T we're currently working.

Edited September 8, 2015 by Sensei

[swansont]

 

Even my hot tea that I just measured has 57.5 C at near top, and 54.5 C at bottom (measured by my -50 C...+300 C thermometer with +-0.1 C precision)

 

But a cup of tea is not a good example. The cup is probably not well-insulated and more importantly it's open to the air at the top. That's why you have a gradient. Put it in a sealed, well-insulated container (like a dewar) and wait a while, and that gradient gets a lot smaller.

[John Cuthber]

 

If volume of water in container is 100 mL = 100 g, 4.1855 J/K*g * 100g * 1 K = 418.55 / 0.05 = 8371 seconds wait for increase water temperature by 1 C..

 

So, if you were sensible and used 10 ml of water and measured the temperature with a hundred year old thermometer like this one

https://en.wikipedia.org/wiki/Beckmann_thermometer

which resolves to 0.001 C you could measure the effect to about 10% relative error in 8 seconds.

 

You don't seem to understand how good modern science is and ,until you do, you have very little chance of coming up with any improvement.