# Rate of Cooling

I was given the following problem:

Two objects have a temperature difference of 20 degrees C at time 0. Two minutes later, they have a temperature difference of 18 degrees. How long does it take for them to have a difference of 10 degrees.

Initially, I tried to approach the problem with Newton's Law of Cooling. However, I believe my solution is flawed:

$$\frac{dT}{dt} = -kT$$

letting $T$ be the temperature difference between the two objects. The solution to the above equation is $T(t) = ce^{-kt}$. Inserting the provided values, I found that $c = T(0) = 20$ and $T(2) = 18 = 20e^{-2k}$, so $k = 0.05268$. Setting $20e^{-0.05268t} = 10$, I found that $t = 13.2$ minutes.

Is Newton's Law of Cooling applicable if I simply let $T$ be the difference in temperatures between two objects? Also, the problem implies that both objects are changing temperature. What do I need to do to accommodate for both objects changing temperatures?

• Excellent example of how to ask a good homework(-like) question :-) – David Z Mar 21 '13 at 0:53

Now, since you asked about temperature difference as opposed to just temperature: Newton's Law is actually about temperature differences, and doesn't treat absolute temperatures. Look at what the equation you posted says: the rate at which $T$ changes is proportional by $-k$ to $T$. The object will stop changing temperature when $dT$ is zero, which will be when $T$ is zero. Would it make sense for something left out in a room to get colder until it reached absolute zero? Not at all! An object left out to the air cools (or warms up) until it's the same temperature as the air around it. So really, the $T$ in your equation is temperature difference between the two objects, not the temperature of just one object. While it's true that both objects will change temperature, the heat exchange between them will always be proportional to their temperature difference, and that difference will approach zero just as you calculated. We might not know what the final temperature will be, but we still know how fast they'll get closer to being the same temperature. Your solution is sound.