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+1 vote
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in Physics by (44.7k points)

A liquid takes 5 minute to cool from 80°C to 50°C. The temperature of the surrounding is 20°C. What is the time it will take to cool from 60°C to 30°C? 

(A) 9 min 

(B) 5 min 

(C) 12 min 

(D) 6 min

1 Answer

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Best answer

The correct option is (A) 9 min.

Explanation:

If a body cools from 80°C to 50°C, average temperature of liquid = [{80 + 50} / 2] = 65°C (T – To) = excess temperature = 65 – 20 = 45°C rate of fall of temperature = {(dθ1) / dt} = {(80 – 50) / 5} = (30 / 5) = 6°C/min (1) 

If body cools from 60°C to 30°C, average temperature of liquid = {(60 + 30) / 2} = 45°C (T – To) = excess temperature = 45 – 20 = 25°C 

Rate of fall of temperature = {(dθ2) / dt} = {(60 – 30) / t} = (30 / t)°C/min (2) 

As {(– dθ) / dt} = K(T – To) ---- Newtons law of cooling 

∴ from (1), – 6 = K(45) 

from (2) {(– 30) / t} = K(25) 

∴ Taking ratio [{– 6} / {(– 30) / t}] = {45 / 25} 

∴ (t/5) = (45 / 25) 

∴ t = {(45 × 5) / 25} 

t = 9 min

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