Experiment
In Science Olympiad there is this competition called Thermodynamics. In thermodynamics, we have to make an insulating box that is smaller than 15cm by 15cm by 15cm. This box will have a 250mL beaker inside of it. The supervisors will heat up water between 60 and 90 degrees Celsius and will be between 50 and 150 milliliters.
I would like to find an equation that will predict the temperature of the water as a function of time, starting temperature, and starting volume.
My Attempt
I used experimental data to try and make an equation for exponential decay because the rate of Heat Loss would slow down. I got that the equation should be in the form:
$a\times \frac{(T_i-T_r)}{w_v}\times b^{ct}+T_r$
In the equation portrayed above $a$, $b$, and $c$ are just constants. $T_i$ is the initial temperature, $T_r$ is room temperature, $w_v$ is water volume, and $t$ is time.
When I made my equation guess, it wasn't accurate, so I don't think I am doing the equation correctly? What equation can I use to calculate the temperature of the beaker over time as a function of time, volume, initial temperature, and room temperature.
Best Answer
Nice intuition! When you say your equation wasn't accurate, do you mean that it didn't accurately fit experimentally obtained time-temperature measurements? That's a little surprising, because I'd expect a similar relationship.
If the insulation is good enough, we might assume that the temperature inside the box and throughout the beaker is uniform (because heat can transfer much more easily within the box than through the sides). In heat transfer analysis, this is called a lumped-capacitance assumption. It lets us ascribe a single temperature $T$ to the inside of the box over time $t$: $T(t)$.
I agree with you that the rate of heat loss can be expected to slow down over time. Certain relevant mechanisms of heat transfer, namely, conduction and convection, depend linearly on the difference between two temperatures. Therefore, we might express the rate of heat loss $q$ from the box (in units of power, or energy per unit time) as $$q=C_1\left[T(t)-T_\infty\right]$$
where $C_1$ a constant representing the material properties and geometry of the insulation and the nature of the dominant heat transfer mechanism and $T_\infty$ is the ambient temperature or room temperature.
(You might even decompose $C_1$ into $C_1=\frac{C_2 C_3 A}{d}$ where $C_2$ is solely the strength of the dominant heat transfer mechanism, $C_3$ is a solely a material property such as the thermal conductivity of the insulation, $A$ is the surface area of the box, and $d$ is its thickness. Here, it makes intuitive sense that the rate of heat transfer scales up with increased surface area and is inversely related to the insulation thickness.)
The rate of heat loss is going to affect the remaining energy in the box (including the beaker) over time. That is, we could write in energy terms that $$C_4 \Delta T(t)=-q\Delta t$$
or
$$C_4 \frac{\Delta T(t)}{\Delta t}=-q=-C_1\left[T(t)-T_\infty\right]$$ where $C_4$ represents some type of thermal capacitance of the beaker and water—that is, a constant that connects their temperature to energy—and $\Delta T(t)$ is the temperature change over some time interval $\Delta t$.
(You might even decompose $C_4$ into $C_4=C_5 C_6+C_7C_8$ where $C_5$ and $C_6$ are the heat capacity and mass, respectively, of the water and and $C_7$ and $C_8$ are the heat capacity and mass, respectively, of the beaker. Maybe this last term is negligible; I don't know. Note that it's essential, though, for the units to agree here and throughout this exercise.)
In differential form, this is $$C_4\frac{dT(t)}{dt}=-C_1\left[T(t)-T_\infty\right]$$ and the solution to this differential equation is $$T(t)=T_\infty+(T_0-T_\infty)\exp(-C_1t/C_4)$$
where $T_0$ is the initial temperature. This equation has a form very similar to the one you postulated. Understanding its derivation may be of use as you continue to characterize and optimize your system.