A body which is initially at 800C cools to 640C in 5 min and to 520C in 10 min.  The temperature of the surrounding should be

Using Newton's law of cooling∆θ∆t = K[θav-θ0](θ1-θ2t) = K[θ1+θ22-θo]In the 1st case, 80-645 = K[80+642-θo]Or 3.2 = K[72-θ0]-------(i)In the 2nd case, 64-525 = K[64-522-θ0]or 2.4 = K(58-θ0) ------(ii)From (i) by (ii), we get 3.22.4 = 72-θ058-θ0Or 185.6-3.2θ0 = 172.8 -2.4θ0 or θ0 = 160C