A body initially at `80^(@)C` cools to `64^(@)C` in 5 minutes and to `52^(@)C` in 10 minutes. What will be its temperature in 15 minutes and what is the temperature of its surroundings?

Explanation

Using average form of Newton's law of cooling, we have
`(80-64)/(5) = k((80+64)/(2)-T_(s))`
`rArr (16)/(5)= k(72 - T_(s))`…(1)
and `(64-52)/(5) = k((64+52)/(2)-T_(s))`
`rArr (12)/(5) = k(58-T_(s))` ….(2)
`((1))/((2))` gives
`(16)/(12) =(72 -T_(s))/()58-T_(s))`
`T=T_(s)+(T_(0)-T_(s))e^(-kt)`
`64 = T_(s)+(80-T_(s))e^(-k(5))`
`8(58-T_(s)) = 6(72 -T_(s))`
`rArr T_(s)=(464 - 432)/(2) = 16^(@)C`
from`((1))`
`(16)/(5) = k(72- 16)`
`rArr k= 0.0571`
Now if in 15 minutes temperature of body is T, we use
`(52-T)/(5) = 0.0571((52 + T)/(2)-16)`
`rArr 104-2T = 14.846 + 0.2855 T - 9.136`
`rarr T=43^(@)C`