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A body cods from $80^{0} c$ to $60^{0} c$ in 2 minutes. In how much time it cools from $60^{0} c$ to $40^{0} c$ ? The temperature of surroundings is 10⁰c ....

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3 min

6 min

2min

1 min

answer is A.

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Detailed Solution

$\frac{d θ}{d t} α (θ_{m e a n} - θ_{s}) \Rightarrow \frac{d θ}{d t_{1}} = - k (\frac{80 + 60}{2} - 10)$

$\frac{d θ}{d t_{2}} = - k (\frac{60 + 40}{2} - 10)$ solving

We get $d t_{2} = 3 \min$

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