A hot body, obeying Newton's law of cooling is cooling down from its peak value 80^ C to an ambient temperature of 30^ C . It takes 5 minutes in cooling down from 80^ C to 40^ C . How much time will it take to cool down

Correct answer: B. 8.6 minutes From Newton's law of cooling, t= (1)/(k) _ e ((θ_2-θ_0)/(θ_1-θ_0)) From question and above equation, 5= (1)/(k) _ e ((40-30))/((80-30)) And, t=(1)/(k) _e ((32-30))/((62-30)) Dividing equation (2) by (1), (t)/(5)=((1)/(k) _e ((32-30))/((62-30)))/((1)/(k) _e ((40-30))/((80-30))) On solving we get, time taken to cool down from 62^ ^ C to 32^ ^ C , t=8.6 minutes.

JEE Main 201411 AprilOnline

Exam: JEE Main
Year: 2014
Date: 11 April
Mode: Online
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2014 · 11 April · Online · Physics · Thermal Properties of Matter

A hot body, obeying Newton's law of cooling is cooling down from...

A hot body, obeying Newton's law of cooling is cooling down from its peak value

80^{\circ} \mathrm{C}

to an ambient temperature of

30^{\circ} \mathrm{C}

. It takes 5 minutes in cooling down from

80^{\circ} \mathrm{C}

40^{\circ} \mathrm{C}

. How much time will it take to cool down from

62^{\circ} \mathrm{C}

32^{\circ} \mathrm{C}

? (Given

\operatorname{In} 2=0.693

, In

5=1.609

)

Options

A $3.75$ minutes
B $8.6$ minutes
C $9.6$ minutes
D $6.5$ minutes

Correct answer

B. $8.6$ minutes

Step-by-step solution

From Newton's law of cooling, t=

\frac{1}{k} \log _

\left(\frac{\theta_2-\theta_0}{\theta_1-\theta_0}\right)

From question and above equation, 5=

\frac{1}{k} \log _

\frac{(40-30)}{(80-30)}

And,

t=\frac{1}{k} \log _e \frac{(32-30)}{(62-30)}

Dividing equation (2) by (1),

\frac{t}{5}=\frac{\frac{1}{k} \log _e \frac{(32-30)}{(62-30)}}{\frac{1}{k} \log _e \frac{(40-30)}{(80-30)}}

On solving we get, time taken to cool down from

62^{{}^{\circ}} \mathrm{C} to 32^{{}^{\circ}} \mathrm{C}

t=8.6

minutes.

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