A hot liquid kept in a beaker cools from 80,^ | vedclass

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(D) According to Newton's law of cooling: $\frac{	heta_{1}-	heta_{2}}{t} = K \left( \frac{	heta_{1}+	heta_{2}}{2} - 	heta_{0} ight)$.For the fi

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$216$

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(D) According to Newton's law of cooling: $\frac{	heta_{1}-	heta_{2}}{t} = K \left( \frac{	heta_{1}+	heta_{2}}{2} - 	heta_{0} ight)$.For the first case: $	heta_{1} = 80\,^{\circ}C$,$	heta_{2} = 70\,^{\circ}C$,$t = 2\, 	ext{min} = 120\, 	ext{s}$,and $	heta_{0} = 30\,^{\circ}C$.$\frac{80-70}{120} = K \left( \frac{80+70}{2} - 30 ight) \Rightarrow \frac{10}{120} = K(75 - 30) \Rightarrow \frac{1}{12} = 45K \Rightarrow K = \frac{1}{540}$.For the second case: $	heta_{1} = 60\,^{\circ}C$,$	heta_{2} = 50\,^{\circ}C$,and $	heta_{0} = 30\,^{\circ}C$.$\frac{60-50}{t} = K \left( \frac{60+50}{2} - 30 ight) \Rightarrow \frac{10}{t} = \frac{1}{540} (55 - 30) \Rightarrow \frac{10}{t} = \frac{25}{540}$.$t = \frac{10 	imes 540}{25} = \frac{5400}{25} = 216\, 	ext{s}$.

$A$ hot liquid kept in a beaker cools from $80\,^{\circ}C$ to $70\,^{\circ}C$ in $2\, \text{min}$. If the surrounding temperature is $30\,^{\circ}C$,then the time taken for the same liquid to cool from $60\,^{\circ}C$ to $50\,^{\circ}C$ is ........ $\text{sec}$.

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