Question 1043870
Suppose that a cup of hot coffee at a temperature of T0 is set down to cool in a room where the temperature is kept at T1. Then the temperature of the coffee as it cools can be modeled by the function . Here, f(t) = (T0 - T1)ekt + T1 is the temperature of the coffee after t minutes; t=0 corresponds to the initial instant when the temperature of the coffee is T0; and k is a (negative) constant that depends, among other factors, on the dimensions of the cup and the material from which it is constructed. [We get this formula in calculus and is based on Newton's law of cooling.] Here are the two problems I want to discuss this week. 
Problem A: Suppose that a cup of hot coffee at a temperature of 185°F is set down to cool in a room where the temperature is kept at 70F°. What is the temperature of the coffee ten minutes later? Use k = -0.15 
Problem B: Suppose that a cup of hot coffee at a temperature of 185°F is set down to cool in a room where the temperature is kept at 70°F. How long will it take for the coffee to cool to 140°F


a.
T0=185,T1=70,K=-0.15,t=time=10 min,

Modelling equation is
{{{f(t) = (T0 - T1)e^(k*t) + T1}}} 
{{{A}}}

find t=10

{{{f(t=10)}}} = {{{(185 - 70)*e^(-0.15*10) + 70}}} 
{{{f(t=10) }}}= {{{(115)*e^(-1.5) + 70}}} 
{{{f(t=10) }}}= {{{(115)*e^(-1.5) + 70}}} 
{{{f(t=10) }}}= {{{(115)*(0.223) + 70}}} 
{{{f(t=10) }}}= {{{25.625 + 70}}} 
{{{f(t=10)}}} = {{{95.625}}}
This means that when time is {{{10}}} min the temperature is {{{95.625}}} which has cooled from 185.


{{{B}}}

{{{140 }}}= {{{(185 - 70)*e^(-0.15*t) + 70}}} 
{{{140 }}}= {{{(115)*e^(-0.15*t) + 70}}} 
{{{70 }}}= {{{(115)*e^(-0.15*t) }}} 
{{{70 }}}= {{{(115)*e^(-0.15*t) }}} 
{{{(70)/(115)}}} = {{{e^(-0.15*t)}}} 
{{{(14/23)}}} = {{{e^(-0.15*t)}}}

Take Ln of both sides
 {{{(Ln(14/23))}}} ={{{lne^(-0.15*t))}}} 

 {{{(Ln(14/23))}}} ={{{(-0.15t*lne)}}} 

{{{ln(e)=1}}}
 {{{(Ln(14/23))}}} ={{{-0.15t*(1)}}} 
 {{{(Ln(14/23))/(-0.15) }}}={{{t }}} 
 {{{(Ln(14/23))/(-0.15)}}} ={{{t }}} 

{{{(-0.496)/(-0.15)}}}={{{t}}}
{{{t=3.3 }}}


time is {{{3.3}}}
it will 3.3 mins to cool to 140.