newtons law of cooling

Question

A cup of coffee at 160° is placed in a room where the temperature is 70°. After 5 minutes, the coffee’s temperature is 112.5°.

 

Find the temperature of the coffee 10 minutes after it was placed in the room.

a. 87.6 degrees
b. 90.1 degrees
c. 95.3 degrees
d. 75 degrees

 

39. After how many minutes in the room will the coffee’s temperature be 85°?

a. 12
b. 15
c. 5
d. 9

Answer 1

Hello, Alex,

 

Here is a solution with answers different from any of the choices you have given.  I post this because on the calculator allowed section of the AP exam you are expected to give answers accurate to three decimal places.  You are also expected to store in memory the best estimate possible of any number you will use in another part of the same question.  I did this with the value of "k" as you will see below.  

 

youYou can begin with the differential equation T′(t)=k·(T(t)-70).  This differential equation is an expression of Newton's Law of Temperature Change (usually called "Newton's Law of Cooling".)  This differential equation is equivalent to

T′(t)/(T(t)-70)=k. By writing expressions for anti-derivatives of each side, we get ln(T(t)-70)=k·t+C.  Then by

using each side as an exponent with base e we get T(t)-70=e^(k·t+C)=e^C·e^k·t.  Since we know T(0)=160 we can determine that e^C = 90.  Then using the fact that T(5)=112.5, we can create the equations 112.5-70=90e^k·5, and 40.5/90=e^k·5.

This means the exact value of k is ln(40.5/90)·1/5.  A very close approximation of this number can be stored in a calculator.  Say this is stored as "A."  Then we can compute T(10) to be very close to 88.225.  

 

To answer the second part of the question, we need a solution of the equation 85=70+90·e^A·t (Remember that our calculator estimate of the value of k is stored as "A". .  This equation is equivalent to the equations 15/90=e^A·t and ln(15/90)·1/A=t.  The solution of this equation is about 11.219 minutes.

Answer 2

By Newton's Law of Cooling, T(t) = T_s + (T₀ - T_s)e^-kt

 

  T(t)=temperature of the cup of coffee t minutes after it is put in the            room

  T_s = temperature of the surroundings

  T₀ = initial temperature of the cup of coffee

 

(a).  Using the given information, we have T(t) = 70 + (160 - 70)e^-kt

 

                                                                  = 70 + 90e^-kt

 

       T(5) = 70 + 90e^-k(5) = 112.5 

 

       So, 90e^-5k = 42.5

 

               e^-5k = 0.47222222

 

                 -5k = ln(0.47222222)

 

                   k = 0.15

 

      Temperature after 10 minutes = T(10) = 70 + 90e^-0.15(10)

 

                                                            ≈ 90.1

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(b). When will the temperature of the coffee reach 85°?

 

      Find t so that T(t) = 85:   70 + 90e^-0.15t = 85

 

                                                  90e^-0.15t = 15

 

                                                     e^-0.15t = 0.16666666

 

                                                      -0.15t = ln(0.166666666)

 

                                                             t ≈ 12 minutes