JEE Main 2022 · 24 Jun · Shift 2 · Physics Q60

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PhysicsInteger+4 / -1Question 60 of 90

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 ______.

Cooling curve of ΔT (°C) against time (minute): ΔT falls from 60 at t = 0, through 40 at t = 6 min and 20 at t = t₂, flattening towards zero; dashed lines mark the points (6, 40) and (t₂, 20).

Your answer

Answer and worked solution

Official answer: 16

Newton's law of cooling (average form): $\dfrac{\Delta(\Delta T)}{t}=k\,\overline{\Delta T}$ .

From 60 to 40 in 6 min. $\dfrac{20}6=k(50)$ , so $k=\dfrac1{15}$ min $^{-1}$ .

From 40 to 20. $\dfrac{20}t=\dfrac1{15}(30)$ , so $t=10$ min and $t_2=6+10=16$ min. (The exponential form gives about 16.3 min, the same to the nearest integer.)

Answer: 16

Topics:Newton's law of cooling

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JEE Main 2022 · 24 Jun · Shift 2, Physics question 60