In an experiment to verify Newton's law of co | vedclass

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(B) According to Newton's law of cooling,the temperature difference $\Delta T$ at time $t$ is given by $\Delta T = \Delta T_{0} e^{-\lambda t}$,where

Vedclass · Accepted Answer

$16$

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(B) According to Newton's law of cooling,the temperature difference $\Delta T$ at time $t$ is given by $\Delta T = \Delta T_{0} e^{-\lambda t}$,where $\Delta T_{0}$ is the initial temperature difference.From the graph,at $t = 0$,$\Delta T_{0} = 60^{\circ}C$.At $t = 6 	ext{ min}$,$\Delta T = 40^{\circ}C$. Substituting these values: $40 = 60 e^{-6\lambda} \Rightarrow e^{-6\lambda} = \frac{40}{60} = \frac{2}{3} \Rightarrow 6\lambda = \ln(1.5)$.At $t = t_{2}$,$\Delta T = 20^{\circ}C$. Substituting these values: $20 = 60 e^{-t_{2}\lambda} \Rightarrow e^{-t_{2}\lambda} = \frac{20}{60} = \frac{1}{3} \Rightarrow t_{2}\lambda = \ln(3)$.Dividing the two equations: $\frac{t_{2}\lambda}{6\lambda} = \frac{\ln(3)}{\ln(1.5)} \Rightarrow \frac{t_{2}}{6} = \frac{1.0986}{0.4055} \approx 2.709$.Therefore,$t_{2} = 6 	imes 2.709 \approx 16.25 	ext{ min}$.Rounding to the nearest integer,$t_{2} \approx 16 	ext{ min}$.

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