A cup of coffee cools from \(90^{\circ}\text{C}\) to \(80^{\circ}\text{C}\) in \(t\) minutes, when the room temperature is \(20^{\circ}\text{C}.\) The time taken by a similar cup of coffee to cool from \(80^{\circ}\text{C}\) to \(60^{\circ}\text{C}\) at room temperature same at \(20^{\circ}\text{C}\) is:
1. \(\frac{10}{13}t\) 
2. \(\frac{5}{13}t\)
3. \(\frac{13}{10}t\)
4. \(\frac{13}{5}t\)

Subtopic:  Newton's Law of Cooling |
 65%
From NCERT
NEET - 2021
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An object kept in a large room having an air temperature of \(25^\circ \text{C}\) takes \(12 ~\text{min}\) to cool from \(80^\circ \text{C}\) to \(70^\circ \text{C}.\) The time taken to cool for the same object from \(70^\circ \text{C}\) to \(60^\circ \text{C}\) would be nearly:
1. \(10 ~\text{min}\)
2. \(12 ~\text{min}\)
3. \(20 ~\text{min}\)
4. \(15 ~\text{min}\)

Subtopic:  Newton's Law of Cooling |
 79%
From NCERT
NEET - 2019
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A body cools from a temperature of \(3T\) to \(2T\) in \(10\) minutes. The room temperature is \(T.\) Assuming that Newton's law of cooling is applicable, the temperature of the body at the end of the next \(10\) minutes will be:

1. \(\frac{7}{4}T\) 2. \(\frac{3}{2}T\)
3. \(\frac{4}{3}T\) 4. \(T\)

Subtopic:  Newton's Law of Cooling |
 73%
From NCERT
NEET - 2016
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A certain quantity of water cools from \(70~^{\circ}\text{C}\) to \(60~^{\circ}\text{C}\) in the first \(5\) minutes and to \(54~^{\circ}\text{C}\) in the next \(5\) minutes. The temperature of the surroundings is:
1. \(45~^{\circ}\text{C}\)
2. \(20~^{\circ}\text{C}\)
3. \(42~^{\circ}\text{C}\)
4. \(10~^{\circ}\text{C}\)

Subtopic:  Newton's Law of Cooling |
 78%
From NCERT
AIPMT - 2014
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