Q. A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80° C to 60°C at a room temperature same at 20°C is:

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

[Solution] Newton's law (average form): 10/t = k(85 - 20) and 20/t' = k(70 - 20). Dividing gives t' = (20 × 65)/(10 × 50) t = 13t/5. Answer (1).

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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