In an experiment to verify Newton's law of cooling, a graph is plotted between the temperature difference $$(\Delta T)$$ 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 ______

To find the value of $$t_2$$, we apply Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the surrounding temperature.

1. Newton's Law of Cooling Formula

The temperature difference $$\Delta T$$ at any time $$t$$ is given by:

$$\Delta T = \Delta T_0 e^{-\lambda t}$$

Where:

• $$\Delta T = T - T_0$$ (Temperature difference at time $$t$$)
• $$\Delta T_0 = T_i - T_0$$ (Initial temperature difference at $$t = 0$$)
• $$\lambda$$ = Cooling constant
• Initial state ($$t = 0$$): $$\Delta T_0 = 80^\circ\text{C} - 20^\circ\text{C} = 60^\circ\text{C}$$ (assuming $$T_0 = 20^\circ\text{C}$$ based on the $$60$$ in the solution).
• At $$t = 6 \text{ min}$$: The graph shows $$\Delta T = 40^\circ\text{C}$$.
• At $$t = t_2$$: The graph shows $$\Delta T = 20^\circ\text{C}$$.

2. Step-by-Step Calculation

Based on the experimental data provided:

$$40 = 60e^{-\lambda(6)}$$

$$\frac{40}{60} = e^{-6\lambda} \implies \frac{2}{3} = e^{-6\lambda}$$

Taking natural log on both sides:

$$6\lambda = \ln\left(\frac{3}{2}\right) = \ln(1.5)$$

$$20 = 60e^{-\lambda t_2}$$

$$\frac{20}{60} = e^{-\lambda t_2} \implies \frac{1}{3} = e^{-\lambda t_2}$$

Taking natural log on both sides:

$$t_2 \lambda = \ln(3)$$

3. Solving for $$t_2$$

Divide the two logarithmic equations to eliminate $$\lambda$$:

$$\frac{t_2 \lambda}{6 \lambda} = \frac{\ln(3)}{\ln(1.5)}$$

$$\frac{t_2}{6} = \frac{\ln(3)}{\ln(1.5)}$$

Using the values $$\ln(3) \approx 1.0986$$ and $$\ln(1.5) \approx 0.4055$$:

$$t_2 = 6 \times \frac{1.0986}{0.4055} \approx 6 \times 2.709$$

$$t_2 \approx 16.25 \text{ min}$$

Final Result:

Rounding to the nearest integer as suggested by the solution:

$$\boxed{t_2 \approx 16}$$