Hot water cools from $80^{\circ} C$ to $60^{\circ} C$ in $1$ minute. How long will it take to cool from $60^{\circ} C$ to $50^{\circ} C$ (in $s$)? (Room temperature $= 30^{\circ} C$)

$A$ hot body,obeying Newton's law of cooling,is cooling down from its peak value $80\,^oC$ to an ambient temperature of $30\,^oC$. It takes $5\,minutes$ to cool down from $80\,^oC$ to $40\,^oC$. How many minutes will it take to cool down from $62\,^oC$ to $32\,^oC$? (Given $\ln 2 = 0.693, \ln 5 = 1.609$)

$A$ body cools down from $52.5^{\circ} C$ to $47.5^{\circ} C$ in $5 \ min$ and from $47.5^{\circ} C$ to $42.5^{\circ} C$ in $7.5 \ min$. Then,the temperature of the surroundings is (in $^{\circ} C$)

$A$ metal rod cools at the rate of $4^{\circ}C/min$ when its temperature is $90^{\circ}C$ and at the rate of $1^{\circ}C/min$ when its temperature is $30^{\circ}C$. The temperature of the surrounding is: (in $^{\circ}C$)

The temperature of a body falls from $62^\circ C$ to $50^\circ C$ in $10$ minutes. If the temperature of the surroundings is $26^\circ C$,the temperature in the next $10$ minutes will become ...... $^\circ C$.

$A$ cup of tea cools from $80\,^{\circ}C$ to $60\,^{\circ}C$ in $1\,min$. The ambient temperature is $30\,^{\circ}C$. In cooling from $60\,^{\circ}C$ to $50\,^{\circ}C$,it will take ....... $sec$.