00:01
Hi, from the question given that a chemical solution at a boiling temperature of 130 degree celsius is set on a table where the room temperature is assumed to be constant at 30 degree celsius.
00:15
The chemical solution is cooled to 55 degree celsius after 5 minutes.
00:20
So, in part if we need to find the expression for the temperature t of the solution after t minutes.
00:24
So, here let t of t will be the temperature of the body after t minutes.
00:39
So, here given that t of 0 is equal to 130 degree celsius.
00:46
Now, the room temperature is t a is equal to 30 degree celsius.
00:55
Now, by newton law of cooling dt by dt is equal to minus rt minus ta, where r is the cooling constant.
01:07
So, dt by dt is equal to minus rt minus 30 that was room temperature.
01:15
Now, by using the variable separable method dt by t minus 30 which is equal to minus r dt.
01:25
Now, ln of t minus 30 which is equal to minus rt plus log c or instead of log we can use here it as natural logarithm ln of c is a integrating constant.
01:42
Now, t minus 30 is equal to ce to the power of minus rt.
01:49
So, from this we obtain t of t is equal to 30 plus 100 sorry 30 plus 100 e to the power of minus rt this 100 can be obtained by substituting t of 0 is equal to 130 degree celsius.
02:09
Since if we substitute substitute the value of t is equal to 0 then 130 is equal to 30 plus ce to the power of r times of 0 is 0...