A body cools from 80 ^o C to 70 ^o C in 5 minutes and further to 60 ^o C in 11 minutes. What will be its temperature after 15 minutes from the start? Also calculate the temperature of the surroundings.

( solve without using log)

The question can be easily solved using the Newton's Law of Cooling, which is

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)........................................(1)

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)........................................(2)

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