Q.
A body initially at 80°C cools to 64°C in 5 minutes and to 52°C in 10 minutes. The temperature of the body after 15 minutes will be
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a
42.7°C
b
35°C
c
47°C
d
40°C
answer is A.
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Detailed Solution
According to Newton's law of cooling, θ1−θ2t=Kθ1+θ22−θ0 For first process: (80−64)5=K80+642−θ0 …(i)For second process: (80−52)10=K80+522−θ0 …(ii)For third process: (80−θ)15=K80+θ2−θ0 …(iii)On solving Eqs. (i) and (ii), we getK=115 and θ0=24∘CPutting these values in Eq. (iii), we getθ=42.7∘C.
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