Predict final temperature by taking temperature samples?

Is it possible to predict what the final temperature will be by taking temperature samples. For example, an object is 0ºC and moved to a room above 0ºC. I'm taking temperature of the object using a thermometer every second. Can I predict (approximation) on what the final temperature would be after a few samples? I guess the more samples the more accurate it would be. Can I calculate when the final temperature might occur based the rate of the temperature change?

Is there any formulas for these kind of calculations?

• Seems certain that a 0 degree body moved into a 25 degree environment will ultimately reach a final temperature of 25 degrees. – Michael Luciuk Feb 4 '13 at 16:49
• The temperature is unknown by the thermometer. – willi Feb 4 '13 at 16:55

Assuming cooling is mainly by convection, the cooling will be described by Newton's law of cooling. This states that the rate of temperature change is proportional to the temperature difference, and result is that the difference between the temperature of your object and the room decays exponentially:

$$T_{room} - T_{object} = (T_{room} - T_{0})e^{-kt}$$

where $T_{0}$ is the initial temperature of your object and $k$ is some constant.

You can take the temperature of the object as a function of time and then fit the expression above, but the problem is that this type of fit gives rather large errors in the final temperature $T_{room}$ unless you measure for long enough that you've almost reached the final temperature. You may also find Newton's law of cooling breaks down when the temperature difference is very small, because there is no longer effective convection.

Let's call the unknown function of temperature with respect to time $Q(t)$. Newton's Law of cooling says that the rate of change of temperate is proportional to the difference in temperature of the object and the environment. I'll write that as a first order linear differential equation, $$\frac{dQ(t)}{dt} = r (T_{f} - Q(t))$$ Where $T_{f}$ is the temperature of the environment, or in our case the unknown final temperate, and $r$ is some unknown constant. Note, when I originally considered this problem I was interested in predicting the final temperate on a thermometer being used to read the temperature of a piece of food that was being cooked. So in this case I am taking that $T_{f} - Q(t) > 0$.

Now let's exine some hypothetical data points. Suppose we take three temperature readings $Q_0,Q_1,Q_2$ at the time intervals $t_0, t_1, t_2$. A requirement here is that the time intervals are equally spaced.

Now lets approximate the rate of change of temperature at time $t=t_1$, by subtracting $Q_1$ from $Q_0$. Putting this into the differential equation gives, $$\frac{Q_1 - Q_0}{t_1 - t_0} = r(T_f - Q_0) \tag{1}$$.

Approximating the rate of change of temperature at $t=t_2$ is given by $$\frac{Q_2 - Q_1}{t_2 - t_1} = r(T_f - Q_1) \tag{2}$$.

Now let's multiply the left hand side of eq (1) by the right hand side of eq (2) and the right hand side of eq (1) by the left hand side of eq (2),

$$\frac{Q_1 - Q_0}{t_1 - t_0}r(T_f - Q_1) = \frac{Q_2 - Q_1}{t_2 - t_1} r(T_f - Q_0)$$,

rewriting,

$$\frac{r}{t_1 - t_0}(Q_1 - Q_0)(T_f - Q_1) = \frac{r}{t_2 - t_1} (Q_2 - Q_1)(T_f - Q_0)$$.

Since we had it a condition that the time intervals are constant we have that $t_1 - t_0 = t_2 - t_1$, thus $\frac{r}{t_1 - t_0} = \frac{r}{t_2 - t_1}$ and thus can be canceled from each side of the eq, giving:

$$(Q_1 - Q_0)(T_f - Q_1) = (Q_2 - Q_1)(T_f - Q_0)$$.

Solving this equation fot $T_f$ gives,

$$T_f = \frac{Q_2 Q_0 - (Q_1)^2}{Q_0 - 2 Q_2 Q_1}$$

We notice that $Q_0 - 2Q_2Q_1 \neq 0$ or $Q_0 \neq 2Q_2Q_1$. But, let's assume that $Q_0 = 2 Q_2 Q_1$. Then, $$\frac{Q_0}{Q_2 Q_1} = 2$$ $$\Leftrightarrow \frac{Q_0}{Q_2}\frac{1}{Q_1} = 2$$

But we have the condition that $0 < Q_0 < Q_1 < Q_2$ since Newtons law of Cooling uses Kelvins for temperature units and we assume that the temperature is increasing across our data samples. Thus $Q_0 < Q_2$ implies $\frac{Q_0 }{Q_2} < 1$. So we must have $Q_1 < \frac{1}{2}$ in order for $Q_0 - 2 Q_2 Q_1$ to equal 0. In fact, if we picked the correct values of $Q_0,Q_1,Q_2$ under 1 we can get a negative value for $T_f$. So, for this model we can require $Q_0,Q_1,Q_2 > 1$ and say that our model breaks down for temperatures below 1 K.

This is a back of the page calculation I did for a side project. I haven't analyzed how much small errors in data samples affect the computation of $T_f$. I assume we can find the same or similar equation for cooling where $Q_2 < Q_1 < Q_0$, but I haven't investigated that yet. Also, we can say that the time intervals are not constant, ie $t_1-t_0 \neg t_2 - t_1$ and instead of canceling those terms, carry them through to the final equation, but I also haven't looked into this yet either.