Hot water cools from 60C  ∘ to 50C  ∘ in the first 10 minutes and to 42°C in the next 10 minutes. The temperature (in C  ∘ ) of the surroundings is:

By Newton's law of cooling ​θ1−θ2t=−K[θ1+θ22−θ0]​ where θ0 is the temperature of surrounding. ​ Now, hot water cools from 60C to   ∘50°C in 10 minutes, ​60−5010=−K[60+502−θ0]......(1)​ Again, it cools from 50°C to 42°C in next 10 minutes.,​50−4210=−K[50+422−θ0]......(2)​ Dividing equations (i) by (ii) we get ​10.8=55−θ046−θ0⇒108=55−θ046−θ0​⇒460−10θ0=440−8θ0​⇒2θ0=20​⇒θ0=10°C