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Question

The temperature of a body falls from $$40^oC$$ to $$36^oC$$ in $$5$$ minutes when placed in a surrounding of constant temperature $$16^oC$$. Then, what is the cooling constant (per minute)?


A
(255)
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B
(115)
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C
loge(56)5
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D
loge(65)5
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Solution

The correct option is C $$\displaystyle - \frac {log_e \big ( \dfrac {5}{6} \big ) }{5} $$
Newton's law of cooling gives us the relation:
$$\dfrac{d\theta}{dt} = -bA(\theta - \theta_{0})$$
Integrating this equation between the appropriate limits, we get:
$$-\dfrac{1}{t} ln(\dfrac{\theta_{2}-\theta_{0}}{\theta_{1} - \theta_{0}}) = bA = $$ Newton's constant
Substituting we get:
Newtons constant $$ = -\dfrac{1}{5} ln(\dfrac{36-16}{40 -16}) = -\dfrac{1}{5}ln(\dfrac{5}{6})$$

Physics

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