Questions
Hot water cools from 60oC to 50oC in the first 10 minutes and to 42oC in the next 10 minutes. The temperature of the surrounding is
detailed solution
Correct option is B
According to Newton's law of cooling, θ1−θ2t=Kθ1+θ22−θ0In the first case, (60−50)10=K60+502−θ0 1=K(55−θ0) ………..(i)In the second case (50−42)10=K50+422−θ0 0.8=K46−θ0 ………(ii)Dividing (i) by (ii), we get 10.8=55−θ46−θor 46−θ0=44−0.8θ0⇒θ0=10∘CSimilar Questions
A body cools down from 60°C to 55°C in 30 s. Using Newton’s law of cooling, calculate the approximate time taken by same body to cool down from 5 5°C
to 50°C. Assume that the temperature of surroundings is 45°C.
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