A cup of coffee cools from $90^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$ in t minutes when the room temperature is $20^{\circ} \mathrm{C}$. The time taken by the similar cup of coffee to cool from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ at the same room temperature is :

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Heat & Thermodynamics

A cup of coffee cools from

9 0^{\circ} C

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

6 0^{\circ} C

at the same room temperature is :

Answer: (A)

Newton's law of cooling in its average-temperature form states

\frac{T _{i} - T _{f}}{t} = k (\frac{T _{i} + T _{f}}{2} - T_{0})

For the first stage, cooling from 90°C to 80°C in time

t

, with room temperature 20°C:

\frac{90 - 80}{t} = k (\frac{90 + 80}{2} - 20) = k (85 - 20) = 65 k ...(i)

For the second stage, cooling from 80°C to 60°C in time

t^{'}

\frac{80 - 60}{t ^{'}} = k (\frac{80 + 60}{2} - 20) = k (70 - 20) = 50 k ...(ii)

Dividing (i) by (ii):

\frac{10/ t}{20/ t ^{'}} = \frac{65}{50} ⟹ \frac{t ^{'}}{2 t} = \frac{13}{10} ⟹ t^{'} = \frac{13}{5} t

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option