A liquid cools from 50°C to 45°C in 5 minutes  and from 45 to 41.5°C in the next 5 minutes. The temperature of the surrounding is

According to Newton's law of cooling msθ1−θ2t∝θ1+θ22−θ0where θ2= temperature of surroundings.In first case,ms50−455∝50+452−θ0                      .....(1)In second case,ms45−41⋅55∝45+41⋅52−θ0               ....(2)Solving eqs, (1) and (2), we getθ0=33⋅3∘C