A body requires 5 minutes to cool from 65^ ^ ^ C to 60^ ^ ^ C . To cool from 60^ ^ ^ C to 55^ ^ ^ C , it will take

Correct answer: C. more than 5 minutes According to Newton's law of cooling, the rate of fall of temperature is given by (Δ T)/(Δ t) = k(T_ avg - T_0) For the first case: (65 - 60)/(5) = k ((65 + 60)/(2) - T_0) 1 = k(62.5 - T_0) For the second case, let the time taken be t : (60 - 55)/(t) = k ((60 + 55)/(2) - T_0) (5)/(t) = k(57.5 - T_0) Since 57.5 - T_0 Thus, (5)/(t) 5 minutes .

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Heat Transfer

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat Transfer

A body requires 5 minutes to cool from 65^ ^ ^ C to 60^ ^ ^ C . To...

A body requires

5

minutes to cool from

65^{}^{{}^{\circ}} C

60^{}^{{}^{\circ}} C

. To cool from

60^{}^{{}^{\circ}} C

55^{}^{{}^{\circ}} C

, it will take

Options

Aless than $5$ minutes
B $5$ minutes
Cmore than $5$ minutes
Deither less or more than $5$ minutes, depending on whether its mass is greater or less than $1 \text{ kg}$

Correct answer

C. more than $5$ minutes

Step-by-step solution

According to Newton's law of cooling, the rate of fall of temperature is given by

\dfrac{\Delta T}{\Delta t} = k(T_{avg} - T_0)

For the first case:

\dfrac{65 - 60}{5} = k \left(\dfrac{65 + 60}{2} -

T_0\right)

<br><br>

1 = k(62.5 - T_0)

<br><br>For the second case, let the time taken be

:<br><br>\dfrac{60 - 55}{t} =

k \left(\dfrac{60 + 55}{2} - T_0\right)

<br><br>

\dfrac{5}{t} = k(57.5 - T_0)

<br><br>Since

57.5 - T_0 < 62.5 - T_0

, it follows that

k(57.5 - T_0) < 1

.<br><br>Thus,

\dfrac{5}{t} < 1 \Rightarrow t > 5 \text{ minutes}

.

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