Calculation of temperature changes (heating a cube in a microwave oven)

[salazar7]

Hello guys,

I am currently running an experiment in my hobby room: I want to heat a cube in a microwave and then measure the surface temperature. Unfortunately, it takes about 20 seconds, so the cube cools down during this time which means I don't measure the actual temperature after heating but the temperature after 20 seconds of cooling.

So let's say the cube has the initial temperature T_0. Now it is heated in the microwave for t_h seconds and now has the temperature T_1. After cooling for t_c seconds, the temperature is measured and is now T_2.

Is it now somehow possible to calculate T_1 using formulas by multiplying here, dividing there, etc.? Perhaps your experience can help me a little. Thank you very much! It doesn't have to be 100% correct at all, so feel free to use assumptions to simplifyPS: I have the following additional data available
Length of the cube (L), specific heat capacity (c_p), thermal conductivity (k), density (R), microwave data: power (P), frequency (f), ambient temperature (T_a).

 

[Chestermiller]

Do you have any idea how to do this?

 

[salazar7]

My very first thought was this:

Assumption: surface temperature presents the average temperature of the whole cube (so the cube is really small/thin)

First, you could calculate the heat absorbed during heating

Q_h = m * c_p * (T_1 - T_0)

Then the cooling

Q_c = m * c_p * (T_2 - T_1)

We could assume that Q_h = Q_c (if this is a realistic assumption)

=> m * c_p * (T_1 - T_0) = m * c_p * (T_2 - T_1) <=> T_1 = (T_2 + T_0) / 2

This somehow doesnt make sense, because we know that T_1 > T_2 (and T_2 > T_0). Let's assume that T_2 = 50°C and T_0 = 20°C. So I get T_1 = 35°C.

 

[Chestermiller]

If there is convective heat loss to the air in the microwave during the heating (and cooling), then the temperature variation with time will not be a straight line.

Let $\Delta T_H(t)$ be the time variation of the surface temperature during an experiment in which the heating is applied. Let $\tau$ be the time at which the microwave is switched off. Since the differential and algebraic equations describing the temperature variation within the system are linear in temperature, the temperature during the cooling period ($t>\tau$) can be expressed as a linear superposition: $$\Delta T(t)=\Delta T_H(t)-\Delta T_H(t-\tau)$$

So, for $t<\tau$, $$\Delta T(t)=\Delta T_H(t)$$ and, for $t>\tau$, $$\Delta T(t)=\Delta T_H(t)-\Delta T_H(t-\tau)$$

 

[salazar7]

thank you very much for your efforts. but unfortunately i'm a bit stumped because i don't know how to implement/apply this. I don't know if this is important to know, but I only have discrete values. So I only know the surface temperature at the time before heating and 20 seconds after heating

 

[russ_watters]

What is the cube made of, how big is it and how are you measuring its surface temperature? Do you have access to an infrared thermometer? That would take about 1 sec after the microwave shuts off.

You can measure the rate of temperature change by measuring the temperature twice and linearly extrapolating backwards. Maybe you'll find the rate is low so this doesn't matter. When the microwave is on you're probably zapping it with 100x more power than rate of loss after it is shut off.

 

[berkeman]

Do you have access to an infrared thermometer?

That was my first thought as well. I use one in the lab pretty often for all kinds of thermal measurements...

https://www.digikey.com/en/products...xnZe5NryTfS5sARzWHqTPeXYGgFIV8TEaAgPDEALw_wcB

 

[Chestermiller]

What are the values of L, thermal conductivity, heat capacity, density, heating time?

 

[salazar7]

thanks for all the advices :)

@Chestermiller I'm currently not at home so i dont know the exact data but maybe we could first work with imaginary values (if thats possible) and when i get home ill try to apply the procedure to the real data:

L = 0,1m
k = 0,5 W/m*K
c_p = 1000 J/kg*K
R = 2000 kg/m^3
t_h = 50 sT_0 = 288,15 K (15°C)
t_c = 100 s
T_2 = 343,15 K (50°C)
P = 1000 Watt
f = 2,45 GHz
T_a = 293,15 K (20°C)

 

[Chestermiller]

thanks for all the advices :)

@Chestermiller I'm currently not at home so i dont know the exact data but maybe we could first work with imaginary values (if thats possible) and when i get home ill try to apply the procedure to the real data:

L = 0,1m
k = 0,5 W/m*K
c_p = 1000 J/kg*K
R = 2000 kg/m^3
t_h = 50 sT_0 = 288,15 K (15°C)
t_c = 100 s
T_2 = 343,15 K (50°C)
P = 1000 Watt
f = 2,45 GHz
T_a = 293,15 K (20°C)

I'll wait to wee the real data. That's what I needed.