Let θ be the temperature of the body at any time t. ∴
dθ
dt
∝(θ−25) ⇒
dθ
dt
=−k(θ−25),k>0 Integrating on both sides, we get log|θ−25|=−kt+c When t=0,θ=80∘ ∴log55=0+c ⇒c=log55 ∴log|θ−25|=−kt+log55.........(i) When t=1 hour =60 minutes, log|θ−25|=