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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Infinity Learn · Accepted Answer

According to Newton's law of cooling, θ$$1$$−θ$$2t=K$$θ$$1+$$θ$$22$$−θ$$0$$ For first process: (80$$−$$64)5=K80+642$$−θ$$0$$           …$$(i)For$$ second process: $$(80$$−$$52)10=K80+522$$−θ$$0$$        …$$(ii)For$$ third process: $$(80$$−θ$$)15=K80+$$θ$$2$$−θ$$0$$                …$$(iii)On$$ solving Eqs. $$(i) and (ii), we $$getK=115$$ and θ$$0=24$$∘CPutting these values in Eq. (iii), we getθ$$=42.7$$∘C.

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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