DIFFERENTIAL EQUATIONS: TEA COOLING QUESTION

THE QUEST to BECOME A MATHEMON MASTER!

Morning Cup of Tea by Flickr user dragonflysky

mornig green tea by Flickr user Kanko*

.:. THE SITUATION .:.

In order to avoid an increased chance of throat cancer for Devon, Jane has to calculate the exact waiting time for the tea to cool down to optimum temperature. In order to do this, Jane, with the help of Derivee, must take down initial values and such to begin working on the question.

.:. THE QUESTION .:.

Determine the time, t, it takes for the tea temperature, T, to cool down to 151°F, given that the initial temperature of the tea was 190°F. After ten minutes, Jane measured the tea to be 167°F. Note that the room’s temperature is 70°F.

.:. THE SOLUTION .:.

The rate is given by

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).

.:. THE SOLUTION .:.

To make things easier, use the separation of variables technique and have the “T”s on one side of the equals sign and the lowercase “t”s on the other.

*k is an unknown constant that will be solved for in future steps.

.:. THE SOLUTION .:.

RESULTING IN:

As you can see, all of the “T”s which represent temperature are on the left and the one dealing with time is on the right.

The next step is to ant differentiate each side.

.:. THE SOLUTION .:.

After anti differentiation we are left with:

REMEMBER that when anti differentiating and equation with out limits, to include the expression of “+C”!!! This is to state that the anti differentiation represents a family of functions.

.:. THE SOLUTION .:.

Also take note that a LOGARITHM is an EXPONENT so that your anti differentiation could be mathematically massaged into the expression:

.:. THE SOLUTION .:.

Considering that there are now three unknown variables, we should try solving for them one at a time. To make this process easier we “eliminate” one variable for now and let

Therefore…

.:. THE SOLUTION .:.

Here, we can solve for A, using the initial value given in the beginning of the question; the initial temperature and the initial time, 0.Solve for A first using initial temperature

Since anything with zero as its exponent is equal to one, A = 120.

.:. THE SOLUTION .:.

Now that we have one constant, we can solve for k using the temperature value at ten minutes, which is 167°F.

 Simplify….

 

.:. THE SOLUTION .:.

To eliminate e, take the ln of both sides and isolate to solve for k. 

**k is approx. -0.0213

.:. THE SOLUTION .:.

Now that we know the previously unknown values, we can now solve for the time it takes to cool to 151°F. The expression looks like this: 

.:. THE SOLUTION .:.

To simplify this complicated expression, find the ln of both sides to take care of the number e. Afterwards, we can then isolate t to find time.

  

t = 18.4717 minutes

SUPER COOLED!!! PROBLEM SOLVED!

It takes approximately 18 minutes in order for the tea to cool down to 151°F

Teapots at the Tea Shop at the Grand Western Canal Country Park in Tiverton by Flickr user pigpogm