A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time ta

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Question

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 the same room temperature is

Accepted Answer

Correct Answer: C — (13/5)tStep-by-step Solution:Using Newton's law of cooling in average form: dθ/dt = -K(θavg - θ₀).Case 1 (90°C to 80°C in time t):(90 - 80)/t = K[(90 + 80)/2 - 20] = K[85 - 20] = 65K → K = 10/(65t) = 2/(13t).Case 2 (80°C to 60°C in time t'):(80 - 60)/t' = K[(80 + 60)/2 - 20] = K[70 - 20] = 50K.20/t' = 50 × 2/(13t) = 100/(13t) → t' = 20 × 13t/100 = 260t/100 = (13/5)t.Why other options are incorrect: Option A (5/13)t is the reciprocal of the correct answer, resulting from inverting the time ratio. Options B and D do not follow from correct application of Newton's law of cooling.Key Concept: Newton's law of cooling (average form): (θ₁ - θ₂)/t = K[(θ₁ + θ₂)/2 - θ₀], where θ₀ is the ambient temperature. This applies when the temperature difference is not very large.

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 the same room temperature is

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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 the same room temperature is

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