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Question

A body at 80oC cools down to 64oC in 5 minutes; and in 10 minutes it cools down to 52oC. What will be its temperature after 20 minutes? What is the temperature of the environment?

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Solution

Using approximate version of Newton's law of cooling
θ1θ2t=α(θ1+θ22θ0)
Now using condition 1:
80645=α[80+642θ0] (1)
Now using condition 2:
645210=α[64+542θ0] (2)
Divide 1 and 2
165×1012=72θ058θ0

83=72θ058θ0

4648θθ=2163θ0

248=5θθ
θθ=2485=49.6°C
Temperature of environment =49.6°C

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