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A body cools down from 45degree C to 40degree C in 5 min and in another 8 min. Find the tem. of the surrounding.?
Verified Answer
A body cools down from 45degree C to 40degree C in 5 min and in anothe...

T0 is the surrounding temperature
in first case
when temperature drops from 80C to 70C in 5 minutes
here, dT = 10C, dt = 5min = 300s
so, we get
-(10/300) = K(80 - T0)
similarly,
in the second case when the temperature drops from 70C to 60C in 11min - 5min = 6 min
here, dT = 10C, dt = 6min = 360s
so,
-(10/360) = K(70 - T0)
dividing (1) by (2) we get
(80 - T0) / (70 - T0) = (10/300) x (360/10) = 6/5
solving further...
400 - 5T0 = 420 - 6T0
or surrounding temperature
T0 = 20C
and 
now, after 15minutes
dt = 15 min = 900 s, dT = (80 - T)
T is the final temperature
so, Newtons Law of Cooling becomes
-(80-T) / 900 = K(80 - 20)
and we also have from 1st case
-10/300 = K(80 - 20)
equating the above two relations. we get
-(80-T) / 900 = -10/300
or
-24000 + 300T = -9000
thus, we have final temperature as
T = 45C
This question is part of UPSC exam. View all NEET courses
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A body cools down from 45degree C to 40degree C in 5 min and in anothe...
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A body cools down from 45degree C to 40degree C in 5 min and in anothe...
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A body cools down from 45degree C to 40degree C in 5 min and in another 8 min. Find the tem. of the surrounding.?
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