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# 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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a
5∘C
b
10∘C
c
15∘C
d
20∘C
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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∘C

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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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