Question 1130119
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Could someone please assist with the question: 
A pot of boiling soup with an internal temperature of 100� Fahrenheit was taken off the stove to cool in a 69�F room.
After fifteen minutes, the internal temperature of the soup was 93�F. 

To the nearest minute, how long will it take the soup to cool to 81�F? 

Here is my work:
69+(100-69) * e= 69+31*e
93=69+31*e(-k*15)
31*e^(-k *15)= 93-69= 24
e^-15k= 24/31= .7742
-15K= ln(.7742)
k= -ln(0.7742)/ 15=0.017

T(t)=  69+31*e (0.017)=81
e(-0.017*t)= 81-69/31=.387
-0.017*t= ln(.387)
t= - ln (.387)/ (0.017)= 55.8 min
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You can use Newton's Law of Cooling.......any of the following 3 formulae should work: {{{highlight_green(system(highlight(matrix(1,3, "T(t)", "=", T[s] + (T[o] - T[s]) * e^(- kt))), OR, highlight(matrix(1,3, dT/dt, "=", - k(T[i] - T[o]))), OR, highlight(matrix(1,3, k(t[1] - t[2]), "=", - ln((T[1] - T[infinity])/(T[2] - T[infinity])))))))}}}

This author's preference is the 1<sup>st</sup> formula. In this case, the cooling rate is first needed, and is derived as follows:

       {{{highlight(highlight_green(highlight(matrix(1,3, "T(t)", "=", T[s] + (T[o] - T[s]) * e^(- kt)) )))}}}, where: {{{t}}} = time taken to get to a COOLED temperature (15 minutes, in this case)
                                             {{{T(t)}}} = TEMPERATURE (T) at a given time (t)___(93<sup>o</sup>F, in this case) 
                                              {{{T[s]}}} = SURROUNDING Temperature (69<sup>o</sup>F, in this case)
                                              {{{T[o]}}} = ORIGINAL/INITIAL temperature (100<sup>o</sup>F, in this case)
                                               {{{k}}} = the CONSTANT or COOLING rate (UNKNOWN, in this case)
                                      {{{highlight(highlight_green(highlight(matrix(1,3, "T(t)", "=", T[s] + (T[o] - T[s]) * e^(- kt)) )))}}}
                                      {{{matrix(1,3, "T(15)", "=", 69 + (100 - 69) * e^(- k*15))}}} ----- Substituting 15 for t, 69<sup>o</sup> for {{{T[s]}}}, and 100<sup>o</sup> for {{{T[o]}}}
                                             {{{matrix(1,3, 93, "=", 69 + (100 - 69) * e^(- 15k))}}} ----- Substituting 93<sup>o</sup> for T(15)
                                             {{{matrix(3,3, 93, "=", 69 + 31e^(- 15k), 24, "=", 31e^(- 15k), 24/31, "=", e^(- 15k))}}}
                                       {{{matrix(1,3, - 15k, "=", ln (24/31))}}} ----- Converting to NATURAL LOGARITHMIC (ln) form
                                              {{{matrix(1,5, k, "=", ln(24/31)/(- 15), "=", 0.0170622)}}}
***********************************************************************
To the nearest minute, how long will it take the soup to cool to 81�F?

                                    {{{highlight(highlight_green(highlight(matrix(1,3, "T(t)", "=", T[s] + (T[o] - T[s]) * e^(- kt)) )))}}}
                                      {{{matrix(1,3, "T(t)", "=", 69 + (100 - 69) * e^(- .0170622t))}}} ----- Substituting 69<sup>o</sup> for {{{T[s]}}}, 100<sup>o</sup> for {{{T[o]}}}, and .0170622 for k
                                          {{{matrix(1,3, 81, "=", 69 + (100 - 69) * e^(- .0170622t))}}} ----- Substituting 81<sup>o</sup> for T(t)
                                          {{{matrix(3,3, 81, "=", 69 + 31e^(- .0170622t), 12, "=", 31e^(- .0170622t), 12/31, "=", e^(- .0170622t))}}}
                        {{{matrix(1,3, - .0170622t, "=", ln (12/31))}}} ----- Converting to NATURAL LOGARITHMIC (ln) form
Time it takes for the soup to cool to 81<sup>o</sup>F, or {{{matrix(1,9, t, "=", ln(12/31)/(- .0170622), "=", "55.624665,", or, 56, minutes, "(approximately)")}}}

As your answer, 55.8 min, or approximately 56 mins, coincides with mine, it is correct!

Great JOB!!</font></font></font></b></pre>