A cup of coffee cools from 90^ C to 80^ C in t minutes when the room temperature is 20^ C. The time taken by the similar cup of coffee to cool from 80^ C to 60^ C at the same room temperature is:

Question

Accepted Answer

Correct answer: D. (d) By using average form of Newton's law of cooling 90-80t=k( 90+802-20) … . (i ) 80-60t^ '=k( 80+602-20) … . ( ii ) (i)/(ii) 10 t^ 't 20= 6550 t^ '= 6550 2t= 6525t= 135t

$A cup of coffee cools from 90^{\circ} C to 80^{\circ} C in t minutes when the room temperature is 20^{\circ} C . The time taken by the similar cup of coffee to cool from 80^{\circ} C to 60^{\circ} C at the same room temperature is:$

Solution

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