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 a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is :

CONCEPT:

 According to Newton's Law of Cooling, we can find out the time taken by the cup, we have;

\(\frac{{{T_1} - {T_2}}}{Δ t} = \ k \left( {\frac{{{T_1} + {T_2}}}{2} - {T_s}} \right)\)

Where T1 and T2 are the initial and final temperature and Ts is the surrounding temperature, and Δt is the time taken and k is a positive constant depending upon the area and nature of the surface of the body.

CALCULATION:

By using Newton's law of cooling, 

\(\frac{{{T_1} - {T_2}}}{\ t} = \ k \left( {\frac{{{T_1} + {T_2}}}{2} - {T_s}} \right)----(1)\)

For the first case we have;

Given: Initial Temperature, \({T_1} = {90^o}C\)

Final temperature, \( {T_2} = {80^o}C \)

and surrounding temperature, \( {T_s} = {20^o}C \)

Now, on putting the above values in equation (1) we get;

\(\frac{{{90} - {80}}}{ t} = \ k \left( {\frac{{{90} + {80}}}{2} - {20}} \right) \\ \Rightarrow k\left[ {85 - 20} \right] = \frac{{90 - 80}}{t}\\ \Rightarrow k\left[ {65} \right] = \frac{{10}}{t} \\ \Rightarrow k = \frac{{2}}{13t} ---(2)\)

For the second case we have;

Similarly, 

Given: Initial Temperature, \({T_1} = {80^o}C\)

Final temperature, \( {T_2} = {60^o}C \)

and surrounding temperature, \( {T_s} = {20^o}C \)

\( \frac{{\left( {80 - 60} \right)}}{{{t'}}}= k\left( {\frac{{80 + 60}}{2} - 20} \right) \\ \Rightarrow k\left[ {50} \right] = \frac{{20}}{t'} ---(3)\)

where t' is the time taken in the second case

substituting the equation (2) in (3) we get;

\( \frac{2}{{13t}}\left( {50} \right) = \frac{{20}}{{{t'}}}\\ ​​ \Rightarrow {t'} = \frac{{13t}}{5}\)

Hence, option 3) is the correct answer.