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

Question

Aicharya · Accepted Answer

According to the Newton's law of cooling, $- \frac{d T}{d t} = K (T - T_{0})$ .

Applying approximation, $- \frac{∆ T}{∆ t} = K (\frac{T_{1} + T_{2}}{2} - T_{0})$

For first $6 \min$ , $T_{1} = 80 ° C, T_{2} = (40 + 20) = 60 ° C, T_{0} = 20 ° C$

$\Rightarrow - \frac{(60 - 80)}{6} = K (70 - 20) \Rightarrow K = - \frac{(60 - 80)}{6 \times (70 - 20)}$

$\Rightarrow K = \frac{20}{6 \times 50} = \frac{1}{15}$

For $6 to t_{2}$ , $T_{1} = 60 ° C, T_{2} = (20 + 20) = 40 ° C, T_{0} = 20 ° C$

$- \frac{(60 - 80)}{(t_{2} - 6)} = K (50 - 20) \Rightarrow t_{2} - 6 = 15 \times \frac{20}{30} \Rightarrow t_{2} = 16 \min$

In an experiment to verify Newton's law of cooling. a graph is plotted between. the temperature…

Solution

In an experiment to verify Newton&#39;s law of cooling. a graph is plotted between. the temperature…

Solution

In an experiment to verify Newton's law of cooling. a graph is plotted between. the temperature…