00:01
The initial temperature that is a initial temperature of the water that is a boiling point at a time t is equal to 0 is given as 100 degrees celsius and the room temperature that is t0 is equal to 20 degrees celsius so by using newton's law of cooling we have d tt over d t is equal to negative k bracket t negative t -nogative t -k and we have negative k t negative 20 so this differential equation can be written it t over t negative 20 is equal to negative k t t on integration both side we have d t over t t negative 20 is equal to negative k of t t t plus c is a constant of interpretation we have a natural log of b negative 20 is equal to negative k t plus c taking log on 20 on both side we have p negative 20 is equal to c to the bar c negative and this it implies p negative 20 is equal to 8 to the bar c that is a multiplied to 8 to negative k t so that our required temperature t of t is equal to a it is the bar negative k t plus 20 but we have t of 0 is equal to 100 degrees celsius it replaced t by 0 t of 0 80 is 100 this is equal to a into the bar 0 that is the 1 plus 20 value of is equal to 100 negative 20 is equal to 80.
02:50
Therefore, the t of t is equal to 80 into the negative a of t plus 20.
03:03
But after the 5 minutes, the temperature of the water becomes 20 become 80 degrees celsius.
03:19
So at t is equal to 5, temperature of the water become 80 degrees celsius.
03:27
So therefore 80 degree, 80 is equal to 80 p .2 .5x0 x x multiplied to k plus 20.
03:42
So from here, 80 multiplied to 0 to 0 .5k is equal to 60.
03:49
It implies it is the per negative 5k is equal to 60 over 80 and this is equal to 3 over 4.
04:02
Now, negative 5k is equal to the natural log of 3 over 4.
04:09
From here, the value of k is equal to 1 over 5, the natural log of 4 over 3.
04:18
And this is equal to 0 .057753.
04:32
Therefore, our temperature of the water, t of the water, t, is equal to 80...