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)
B
(115)
C
loge(56)5
D
loge(65)5

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 constantSubstituting 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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