Area under the curve of a temperature-time graph -> energy?

[salazar7]

Hello everyone, hope you are all well. I have the following problem:

I have a temperatur-time graph. If you determine the integral of this graph, you get the unit [kelvin*second]. This unit is as far as I know meaningless.
Is it possible to mathematically "transform" the area under the curve in such a way that [joule] or [watt] comes out or something like that - perhaps a unit which i can interpret, maybe you have some ideas.

Thank you very much for your time!

 

[Lnewqban]

Welcome, Salazar!

If the temperature is changing with time, other things need to be considered before talking about energy gained or lost by certain mass of substance.
Please, see:
https://en.wikipedia.org/wiki/Heat_capacity

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

 

[salazar7]

Thanks Lnewqban for the warm welcome and the quick reply. You're absolutely right. Perhaps it will help if I explain the exerpiment:

I have a test specimen (cuboid) that has a hole in it. A temperature sensor is inserted into this hole to measure the heating of the test specimen from inside over time. Let's say the heating takes 10 seconds. The pressure is constant, so we can work with c_p and ignore the fact that at higher temperatures it takes more energy to change the temperature. I am only interested in the temperature inside the specimen, so we only focus on heat conduction (because "heat conduction occurs as rapidly moving atoms interact with neighboring atoms transferring some of their energy to these neighboring particels" and I think thats what i am looking for)

As an example, if you want to calculate with values, we could take the following values:

• f(t) = -0,3t^2 + 5t + 13
• Dimensions of the body: L = 0,1m, b = 0,1m, d = 0,1m
• c_p = 1000 J/kg*K
• k = 0,1 W/m*K
• Rho = 2000 kg/m^3

I hope that was what you wanted to hear. If not, please feel free to ask further questions

Thanks!

 

[Chestermiller]

What is the cube being heated by? Where is the heat source located? You have a cube of material 0.1 m cube, right? Where is the thermocouple located within the cube (geometrically), relative to the heat source?

 

[salazar7]

The cube is heated using a microwave "oven". Unfortunately, I cant tell you the exact distance to the heat source, but I hope that it probably doesn't matter any more (correct me if I'm wrong), as the heat is "generated" in the specimen itself because of the heating mechanism of the microwaves. The temperatur sensor is placed right in the middle of the cube

 

[Chestermiller]

Is the thermocouple passive or does it directly absorb microwave energy?

If the microwave energy is generated uniformly throughout the cube and negligible energy escapes from the cube (to the air in the microwave) during the heating period, the change in internal energy of the cube is $mC_p\Delta T$, where m is the mass of the cube. During the heating, the thermocouple temperature should vary linearly with time.

 

[salazar7]

And the DeltaT would be T(10) - T(0)? i.e. the temperature at the end of heating minus the temperatur before the start of heating right?

But can you think of any way to interpret the integral of a temperatur-time graph, even if we have to multipy/divide it by something beforehand? Or should I really just stick with "m*C_p*DeltaT" for the evaluation?

thank you very much for your time!

 

[Mister T]

If find the area under the temperature-time graph (by taking the integral, presumably using numerical methods) and then divide that by the total time, that will give you the average temperature.

 

[Steve4Physics]

As an example, if you want to calculate with values, we could take the following values:

• f(t) = -0,3t^2 + 5t + 13
• Dimensions of the body: L = 0,1m, b = 0,1m, d = 0,1m
• c_p = 1000 J/kg*K
• k = 0,1 W/m*K
• Rho = 2000 kg/m^3

The experimental details aren’t clear but your example above for f(t) suggests something is being heated up and then cooling down during the measurements.

If so, it may be useful to use the slopes of the graph at different times.

1. Construct the tangent to the curve at a chosen time.

2. Determine the slope (gradient of the tangent); I’ll represent it as $\frac {dT}{dt}$ (assuming you are familiar with calculus notation) where $T$ is temperature (in ºC or K) and t is time (in seconds). The slope is has units of K/s.

3. If you know the cube's heat capacity, $C$ (in J/K), then $C\frac {dT}{dt}$ corresponds to the net instantaneous power (in watts) transferred to/from the cube. Warning: this may be very inaccurate because the temperature of the sensor could be different to the temperature of other parts of the cube.

If you wanted, you could find the instantaneous power at a number of different times (repeat steps 1-3). Then plot a new graph of power vs time. The appropriate area of this graph gives the net energy-transfer - but the value could be very inaccurate as explained above.

 

[sophiecentaur]

Is it possible to mathematically "transform" the area under the curve

During the heating, the thermocouple temperature should vary linearly with time.

If the OP has observed a "curve" on his results graph then there must be some significant heat loss from the surface of the cube. I remember plotting 'cooling curves' in my school lab to estimate the rate of heat loss during heating experiments. Can we see his data please?

To get a curve in the heating plot would imply that the cube was getting pretty hot by the end of the heating process. A cooling curve could be revealing here. The terminal temperature inside a 1kW oven could be 'dangerously' high for a kitchen device. I remember looking at adverts for small, cheap enamelling ovens which are claimed to run inside kitchen microwave ovens. Enamelling involves dull red heat temperatures so I wonder how well they work or if the glass oven trays survive.

 

[salazar7]

@Steve4Physics oh that was my mistake. I actually only wanted to show the function of the heating process but I accidentally let the temperature sensor continue to run after the heating process and forgot to shorten the diagram on the time axis, which is why a quadratic function is displayed and not a linear one for example.

but anyway, thanks for all your help, I really appreciate it. I'll start with the evaluation and see if something good comes out of it.