Temperature

• Dec 6th 2009, 04:19 PM
Morgan82
Temperature
A pan of water (46 degrees) was put into a fridge. Ten min later, its temp was 39. Ten min after that, it was 33. Use newtons law of cooling. How cold was the fridge?

I know his law, but how do I find the rate?
• Dec 6th 2009, 06:50 PM
HallsofIvy
Quote:

Originally Posted by Morgan82
A pan of water (46 degrees) was put into a fridge. Ten min later, its temp was 39. Ten min after that, it was 33. Use newtons law of cooling. How cold was the fridge?

I know his law, but how do I find the rate?

Let $\displaystyle T_R$ be the temperature inside the refridgerator (which we can assume is kept constant) and let T(t) be the temperature of the water after t minutes. "Newton's law of cooling says that heat will flow from a hotter object to a cooler at a rate proportional to the difference in temperature. Since the temperature is proportional to the amount of heat in the object, we can put the two proportions together to say $\displaystyle \frac{dT}{dt}= k (T- T_R)$ where "k" is the, yet unknown, constant of proportion.

That is a relatively simple separable differential equation which you can solve by integrating both sides of $\displaystyle \frac{dT}{T- T_R}= k dt$.

That will, of course, depend upon three unknown constants, k, $\displaystyle T_R$, and the constant of integration. Fortunately, you have three equations, T(0)= 46, T(10)= 39, and T(20)= 33. You can solve those three equations for the three constants, in particular $\displaystyle T_R$ which is what you are asked.