Question
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
Answer: Option A
:
A
According to Newton law of cooling
θ1−θ2t=K[θ1−θ22−θ0]
For first process : (80−64)5=K[80+642−θ0] ....(i)
For second process : (80−52)10=K[80+522−θ0] ....(ii)
For third process : (80−θ)15=K[80+θ2−θ0] ....(iii)
On solving equation (i) and (ii) we get K=115 and θ0=24∘C . Putting these values in equation (iii) we get θ=42.7∘C
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:
A
According to Newton law of cooling
θ1−θ2t=K[θ1−θ22−θ0]
For first process : (80−64)5=K[80+642−θ0] ....(i)
For second process : (80−52)10=K[80+522−θ0] ....(ii)
For third process : (80−θ)15=K[80+θ2−θ0] ....(iii)
On solving equation (i) and (ii) we get K=115 and θ0=24∘C . Putting these values in equation (iii) we get θ=42.7∘C
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