00:01
Hello students in this question we have a liquid which take time t1 5 minute from temperature t1 equals to 80 degrees centigrade to come out to t2 equals to 50 degrees centigrade.
00:14
So how much time it will take t2 from to cool from temperature t3 equals to 60 degrees centigrade to t4 equals to 30 degrees centigrade and we have given the temperature of the surrounding it is equals to 20 degrees centigrade.
00:30
So from the newton's law, we can write that the temperature t1 minus t2 divide by time t, this is equals to constant k and average temperature that is t1 plus t2 divided by two minus temperature of the surrounding.
00:46
So first, for this data, we can write that t1 which is 80, t2 which is 50, and time t1 is 5 minute, this is equal to k, and t1 is 80.
00:58
T2 is 50 divided by 2 minus surrounding temperature is 20.
01:02
Okay.
01:03
So from here we get this value will be equals to 30.
01:08
So 30 divided by 5 will be equals to 6.
01:10
So 6 is equal to k...