JEE Main 2025 Jan 29 Shift 2, Physics Q19: Newton’s Law of Cooling

JEE Main2025Jan 29, Shift 2Physics

A cup of coffee cools from $9 0^{\circ}$ C to $8 0^{\circ}$ C in $t$ minutes when the room temperature is $2 0^{\circ}$ C. The time taken by the similar cup of coffee to cool from $8 0^{\circ}$ C to $6 0^{\circ}$ C at the same room temperature is :

A
$\frac{13}{5} t$
B
$\frac{10}{13} t$
C
$\frac{13}{10} t$
D
$\frac{5}{13} t$

Solution

Apply the “mean-temperature” form of Newton's law of cooling, $\frac{T _{1} - T _{2}}{Δ t} = k (\frac{T _{1} + T _{2}}{2} - T_{s})$ .

Stage 1: $\frac{90 - 80}{t} = k (\frac{90 + 80}{2} - 20) = 65 k$ , so $\frac{10}{t} = 65 k$ .

Stage 2: $\frac{80 - 60}{t ^{'}} = k (\frac{80 + 60}{2} - 20) = 50 k$ , so $\frac{20}{t ^{'}} = 50 k$ .

Take the ratio: $\frac{10/ t}{20/ t ^{'}} = \frac{65}{50} \Rightarrow \frac{t ^{'}}{2 t} = \frac{13}{10} \Rightarrow t^{'} = \frac{13}{5} t$ .

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JEE Main 2025 Jan 29 Shift 2, Physics Q19: Newton’s Law of Cooling

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