A liquid cools from 500C to 450C in 5 min and from 450C to 41.50C in the next 5 min. The temperature of the surrounding is

Applying θ1−θ2t=αθ1+θ22−θ0, we get  (where, α is constant) 50−455=α50+452−θ0             …(i)and   45−41.55=α45+41.52−θ0                 …(ii)Solving these two equations, we getθ0 = temperature of surrounding = 33.30C.