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Q.

In an experiment to verify Newton's law of cooling, a graph is plotted between, the temperature difference $(Δ T)$ of the water and surroundings and time as shown in figure. The initial temperature of the water is taken as $80 ° C$ . The value of $t_{2}$ as mentioned in the graph will be

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answer is 16.

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Detailed Solution

The initial temperature of the water is taken as $80 ° C$ . The value of $t_{2}$ as mentioned in the graph will be 16 min.

$T - T_{0} (T_{i} - T_{0}) e^{- \frac{Bt}{ms}}$
$6 λ = \ln 1.5$
$40 = 60 e^{- λ (6)} \Rightarrow 6 λ = \ln 1.5$
$20 = 60 e^{- λ t_{2}} \Rightarrow t_{2} λ = \ln 3$
$\frac{t_{2}}{6} = \frac{\ln 3}{\ln 1.5}$
$∴ t_{2} = 16.25 \min$
$So \approx 16$

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In an experiment to verify Newton's law of cooling, a graph is plotted between, the temperature difference (ΔT) of the water and surroundings and time as shown in figure. The initial temperature of the water is taken as 80°C. The value of t2 as mentioned in the graph will be