The soup had just simmered reaching S degrees C, but the fridge was not powerful

Question

The soup had just simmered reaching S degrees C, but the fridge was not powerful enough to accomodate a big pot of soup if it was any warmer than F 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 W degrees C) and stirring occasionally, he could bring the temperature of the soup to P degrees in m minutes.

How long before closing time should the soup be ready so that Jim could put it in the fridge and leave on time?

Parameters:
S= simmering temp of the soup    F= maximum temp that the soup can be put in the fridge    W= temp of the water in the sink where the pot is cooled    P= temp of the cooled soup after m minutes in the water of temp W    m= the number of minutes the pot stays in the sink of water of temp W

Values of parameters: S= 100    F= 15     W= 9   P=60     m=8

Use notation Q(t) for the temp of the soup at any time t.

Explanation / Answer

Using newtons equation of cooling we have the expression ln([T(t)-Tm]/[T(0)-Tm]) = -kt where T is the temperature of system at any time t and k is the rate constant for cooling We have the information about the rate of cooling at a particular time t = 8 min . We need to find the time when the soup reaches temperature T = 15 degrees when kept in the cold water of temp 9 degress we know ln(60 -9/(100-9)) = -k* 8 Time when the soup reaches 15 degress will be ln (15 - 9/100-9) = -k*t dividing the two expression removes the unknown k ln(51/91)/ln(6/91) = 8/t t = 8*ln(6/91)/ln(51/91) t = 37.5674 minutes

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