Question 1188251
.
A pan of water is brought to a boil and then removed from the heat. 
Every 5 minutes thereafter the difference between the temperature 
of the water and room temperature is reduced by 50%.
a) Room temperature is 20 degrees Celsius. Express the temperature of the water 
as a function of the time since it was removed from the heat.
b) How many minutes does it take for the temperature of the water to reach 30 degree Celsius.
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            It is a  STANDARD  problem on the  Newton's law of cooling,

            which is a standard application of logarithms and exponential functions.


            So,  it is  TOTALLY  GOOD  ( legal )  for this forum.



<pre>
I will assume that the boiling temperature is 100 °C (standard for water at normal conditions).


Then according to the Newton's cooling law, the temperature of the water at any time moment t 
after removing from the heat is


    T(t) = 20 + {{{(100-20)*(1/2)^(t/5)}}}.


This formula is the answer to your first question (a).



To answer your second question, write this equation as you read the problem


    30 = {{{20 + 80*(1/2)^(t/5)}}}.


Simplify it step by step


    30 - 10 = {{{80*(1/2)^(t/5)}}}

      20    = {{{80*(1/2)^(t/5)}}}

      {{{20/80}}} = {{{(1/2)^(t/5)}}}

      {{{1/4}}} = {{{(1/2)^(t/5)}}}.


At this point, the answer is OBVIOUS:  {{{t/5}}} = 2,  or  t = 2*5 = 10 minutes.


It is the answer to your second question (b).
</pre>

Solved and thoroughly explained in all details.


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To see many other similar &nbsp;(and different) &nbsp;cooling problems, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/logarithm/Solving-problem-on-Newton-Law-of-cooling.lesson>Solving problem on Newton Law of cooling</A> 

in this site.