Newton's Law of Cooling states that the rate of change of the
temperature of an object is proportional to the difference between its
own temperature and the ambient temperature (i.e. the temperature of
its surroundings).
Newton's Law makes a statement about an instantaneous rate of change of the temperature. We will see that when we translate this verbal statement into a differential equation, we arrive at a differential equation. The solution to this equation will then be a function that tracks the complete record of the temperature over time. Newton's Law would enable us to solve the following problem.
Example 1: The Big Pot of Soup As part of his summer job at a restaurant, Jim learned to cook up a big pot of soup late at night, just before closing time, so that there would be plenty of soup to feed customers the next day. He also found out that, while refrigeration was essential to preserve the soup overnight, the soup was too hot to be put directly into the fridge when it was ready. (The soup had just boiled at 100 degrees C, and the fridge was not powerful enough to accommodate a big pot of soup if it was any warmer than 20 degrees C). Jim discovered that by cooling the pot in a sink full of cold water, (kept running, so that its temperature was roughly constant at 5 degrees C) and stirring occasionally, he could bring that temperature of the soup to 60 degrees C in ten minutes. How long before closing time should the soup be ready so that Jim could put it in the fridge and leave on time ?
Solution: Let us summarize the information briefly and define notation for this problem.
Let
Given: The rate of change of the temperature
, is (by Newton's Law of Cooling equation) proportional to the difference between the temperature of the soup 
and the ambient temperature 
This means that:
![\[ \frac{dT}{dt} {\rm~ is~ proportional~ to~} (T - T_a). \]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_spPhh8Q8TcQO3R_KffOjETwI4r3C1VvcgWansXwAy3disErRIq06K4zmah7MjU6s2512feKOB6RD5TiQGBruWJS_OHxJYsbouDiWxda9_JkuFxpzmYpAru5mrSNPLBTueY51YNnPSwamr1AoDdrKHVOWND1w=s0-d)
Here a bit of care is needed: Clearly if the soup is hotter than the water in the sink
, then the soup is cooling down which means that the derivative 
should be negative.
(Remember the connection between a decreasing function and the sign
of the derivative ?). This means that the equation we need has to have
the following sign pattern:
![\[ \frac{dT}{dt} = -k (T - T_a). \]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ssuQpSWLkJasQGhoXam4Hg80CWE8ejzV_WXeQnMGeQuaRfmLzjPVfyGAO-iwsgFoi9KgZW1EI18PfFppsIWu-9gH4yXZQVk51wtsOmEaAA8x8WQwRbiQeecq8Hw-aL8H43kPsxTxInu2SGW_ORQe8XRqZdFw=s0-d)
where
is a positive constant.
This equation is another example of a differential equation. The independent variable is
for time, the function we want to find is 
, and the quantities 
are constants. In fact, from Jim's measurements, we know that 
, but we still don't know what value to put in for the constant 
.
Newton's Law makes a statement about an instantaneous rate of change of the temperature. We will see that when we translate this verbal statement into a differential equation, we arrive at a differential equation. The solution to this equation will then be a function that tracks the complete record of the temperature over time. Newton's Law would enable us to solve the following problem.
Example 1: The Big Pot of Soup As part of his summer job at a restaurant, Jim learned to cook up a big pot of soup late at night, just before closing time, so that there would be plenty of soup to feed customers the next day. He also found out that, while refrigeration was essential to preserve the soup overnight, the soup was too hot to be put directly into the fridge when it was ready. (The soup had just boiled at 100 degrees C, and the fridge was not powerful enough to accommodate a big pot of soup if it was any warmer than 20 degrees C). Jim discovered that by cooling the pot in a sink full of cold water, (kept running, so that its temperature was roughly constant at 5 degrees C) and stirring occasionally, he could bring that temperature of the soup to 60 degrees C in ten minutes. How long before closing time should the soup be ready so that Jim could put it in the fridge and leave on time ?
Solution: Let us summarize the information briefly and define notation for this problem.
Let
 = Temperature of the soup at time t (in min). |
 = Initial Temperature of the soup =100 deg. |
 = Ambient temperature (temp of water in sink) = 5 deg . |
Given: The rate of change of the temperature
, is (by Newton's Law of Cooling equation) proportional to the difference between the temperature of the soup and the ambient temperature This means that:Here a bit of care is needed: Clearly if the soup is hotter than the water in the sink
, then the soup is cooling down which means that the derivative should be negative.
(Remember the connection between a decreasing function and the sign
of the derivative ?). This means that the equation we need has to have
the following sign pattern:where
is a positive constant. This equation is another example of a differential equation. The independent variable is
for time, the function we want to find is , and the quantities are constants. In fact, from Jim's measurements, we know that , but we still don't know what value to put in for the constant . 
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