00:01
In this question we have been given the newton's law of cooling that states that the rate of cooling of an object is directly proportional to the temperature difference between the object and its surroundings and the equation is d t over d t is equal to k multiplied by t minus a where k is the constant of proportionality t is the temperature of the object and is the temperature of ambient temperature of the object's surroundings.
00:32
So we have to find that if a coffee marks initial temperature we have initial temperature of coffee cup is equal to 170 degrees and after 12 minutes temperature becomes 163 days.
01:14
Celsius celsius we have and the ambient temperature of the room in which the coffee mug is placed is 67 degrees celsius.
01:34
So we have to find how long will it take before the coffee mug reaches 150 degrees.
01:40
So first of all we have the equation d capital t over d small t is equal to k times t minus a so if we write it as d it is if we write it as d capital t over capital t minus a is is equal to k d t and integrate both sides so we have d capital t over t minus a is equal to k integral d t because k is constant so we have not taken it into the integration so we have here ln mod of t minus a will be equal to k t plus c we have k t plus c so here we have so here we can write it as t minus a is equal to e to the bar k t plus c so here we can write t minus a is equal to e to the bar k t multiplied by e to the part c and if we let e to the other constant c dash so we have t minus a is equal to c dash e to the bar k t and final equation is t is equal to c dash e to the bar k t plus a so this is our equation one so this is the now initially the equation was d capital t over d small t is equal to k times capital t minus a where capital t was the temperature of the object after t units of time so we were given that after in 12 minutes the change was that is d capital t over d small t was 170 minus 163 which is equal to k times temperature after in order of time was that is after 12 minutes of 163 minus 67 so the value of k is 0 .073 the value of k is 0 .073 the value of k is 0 .073 so our equation becomes the temperature of the object after time t is equal to c dash e to the bar 0 .073 t plus a so this is a second equation now we have to find out the time t after which the temperature of the cup reaches 150 degrees so after let after time t the temperature of the cup reaches 150 so we have 150 over here is equal to we have and before that we will take out the value c dash so after time 12 minutes we have c dash 8 .0 .073 multiplied by 12 plus 67.
04:59
So here we have, so after 12 minutes the temperature of the cup was 163 is equal to c dash it is the bar 0 .073 multiplied by 12 plus 67.
05:15
So from here we get c dash equal to.
05:19
So we have the value of c approximately as 40.
05:23
So our equation becomes 40 times our equation becomes 40 times e to the power 0 .073 t plus a.
05:39
This is our third equation.
05:41
Now we have to tell the time t after which the temperature reaches 155.
05:47
So we have 150 over here is equal to 40 multiplied by e to the power 0 .073 t plus the ambient temperature 67.
06:00
So we have 150 minus 67 over 40 is equal to e to the bar 0 .073 t...