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  •  A hot body, obeying Newton’s law of cooling is cooling down from its peak value 800C to an ambient temperature of 300C.It takes 5 minutes in cooling down from 80<s

 A hot body, obeying Newton’s law of cooling is cooling down from its peak value 800C to an ambient temperature of 300C.It takes 5 minutes in cooling down from 800C to 400C. How much time will it take to cool down from 620C to 320C ?

(Given ln 2=0.693, ln 5=1.609)

Option: 1

3.75 minutes


Option: 2

8.6 minutes


Option: 3

9.6 minutes


Option: 4

6.5 minutes


Answers (1)

best_answer

 According to Newton's Law of Cooling, if the temperature difference between the body and its surrounding is very small then the Rate of cooling is directly proportional to the temperature difference between the body and its surrounding.

I.e       \frac{d\theta}{dt}\alpha(\theta-\theta_{0})   

Using this we get

$\left(\theta_{\mathrm{t}}-\theta_{0}\right)=\left(\theta_{\mathrm{i}}-\theta_{0}\right) \mathrm{e}^{-\mathrm{kt}}$

where$\theta_{\mathrm{t}}   the temperature at the time t, \theta_{0} the temperature of the surroundings, $\theta_{\mathrm{i}} the peak/initial temperature and k is a constant.

So from the question we have

For cooling from 800C to 400C

 (40-30)=(80-30)e^{-k(5 )}\\ \frac{10}{50}=e^{-k(5 )}\\ ln(5)=5k\\ k=\frac{ln(5)}{5}

Now for  cooling from  620C to 320C 

(32-30)=(62-30)e^{-k(t)}\\ \frac{2}{32}=\frac{1}{16}=e^{-kt} \\ kt=4*ln(2)\\ t=\frac{4*ln(2)}{\frac{ln(5)}{5}}=20*\frac{ln(2)}{ln(5)}=20*\frac{0.693}{1.609}=8.6 \ min

Posted by

SANGALDEEP SINGH

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