A cup of coffee cools from 90^ to 80^ in t minutes, when the room temperature is 20^ . The time taken by a similar cup of coffee to cool from 80^ to 6 - Physics MCQ NEET 2021 2026 | ExamTest.live

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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 a similar cup of coffee to cool from 8 0^{\circ} C to 6 0^{\circ} C at a room temperature same at 2 0^{\circ} C is:

Solution & Step-by-step Explanation

According to Newton's law of cooling: \frac{T _{1} - T _{2}}{t} = K [\frac{T _{1} + T _{2}}{2} - T_{0}] Case 1: \frac{90 - 80}{t} = K [\frac{90 + 80}{2} - 20] \frac{10}{t} = K [85 - 20] = 65 K \dots (i) Case 2: \frac{80 - 60}{t ^{'}} = K [\frac{80 + 60}{2} - 20] \frac{20}{t ^{'}} = K [70 - 20] = 50 K \dots (ii) Dividing (i) by (ii) : \frac{10/ t}{20/ t ^{'}} = \frac{65 K}{50 K} \frac{t ^{'}}{2 t} = \frac{13}{10} t^{'} = \frac{13}{5} t

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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 a similar cup of coffee to cool from 8 0^{\circ} C to 6 0^{\circ} C at a room temperature same at 2 0^{\circ} C is:

\frac{10}{13} t

\frac{5}{13} t

\frac{13}{10} t

\frac{13}{5} t

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