00:01
Hello in the question it is given one object cools down from 80 degrees celsius to 50 degrees celsius and time taken for this scenario that is from 80 degrees celsius to 50 degrees celsius to cool down it will take five minutes time and we need to calculate the time taken to cool down from for the same object from 60 degrees celsius to 30 degrees celsius the time taken that we need to calculate so according to newton's law of cooling we have minus d t divided by d t is equal to k into t minus t not so making the temperature turned together and time in another that is d t divided by t minus t not that is constant we must consider k that is equal to minus k d t consider this an equation one here t is the temperature of the body t not is the temperature of the surrounding and it is given in the question temperature of the surrounding is 20 degrees celsius and k is the constant and small letter t represent the time so temperature of the body falls from 80 degrees celsius to 50 degrees celsius in time 5 minutes so 5 minutes means when we convert this 1 into seconds it will be equal to 300 seconds so integrating this equation 1 so integrating d t divided by k t minus t0 is equal to integration of minus k d t so here this is the temperature term in 300 seconds the object cools down from 80 degrees celsius to 50 degrees celsius so here in temperature term initial value is 50 degree and final value is 80 degree and in time it is 0 to 300 seconds so integrating we will get so one divided by t minus t knot that is log e t minus t0 and from initial 50 to 80 that is equal to minus k and integration of d t is t from 0 to 300 so here converting log e to log 10 that is 2 .303 into log 10 log to the base 10 and here temperature t minus t that is 80 minus t not is 20 divided by 50 minus 20 minus k into 300 minus 0 and when we take k into lhs that is 2 .303 log to the base 10 here divided by k that is 80 minus 20 divided by 50 that is 2 that is equal to 300 minus 300 so here taking 300 to the lhs and k to the rhs we will get 2 .303 divided by minus 300 log to the base 10 2 that is equal to k so consider this as equation 2.
04:06
So, writing this similar equation for the temperature, 60 degrees celsius to 30 degrees celsius, the time will take is t -dash.
04:16
T -dash value we need to calculate.
04:19
So, substituting the values minus 2 .303, log to the base 10 divided by small letter t into here.
04:29
In the place of 80 and 50, we need to substitute 60 and 30.
04:33
So 60 minus 20 divided by 30 minus 20 that is equal to, okay.
04:43
So here minus 2 .3303 divided by t dash log to the base 10 and this value calculating we will get 4 that is equal to k.
04:56
So consider this equation as 3.
04:58
So when we look into equation 2 and 3, the rhs side is equal.
05:02
So we can equate equation 2 and 3...