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
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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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