An object kept in a large room having air temperature of 25°C takes 12 minutes to cool from 80°C to 70°C.

The time taken to cool for the same object from 70°C to 60°C would be nearly,

CONCEPT:

Newton's cooling is written as;

\(\frac{T_1 - T_2}{t} = K (\frac{T_1 + T_2}{2} - T_s)\)

Here we have T₁ is temperature 1,  T1 is temperature 2, K is the positive constant and Ts is the temperature of the surroundings.

CALCULATION:

Given: Temperature  T1  = 80°C

Temperature  T2  = 70°C

time, t = 12 min.

and the temperature of the surroundings, Ts =  25°C

As we know,

\(\frac{T_1 - T_2}{t} = K (\frac{T_1 + T_2}{2} - T_s)\)

Now, on putting the given values we have;

\(\frac{80 - 70}{t} = K (\frac{80 + 70}{2} - 25)\)

\(\Rightarrow \frac{10}{12} = K (\frac{150}{2} - 25)\)

\(\Rightarrow \frac{10}{12} = K (75 - 25)\)   ----(1)

\(\Rightarrow \frac {10}{12} = 50K\)

\(\Rightarrow \frac {1}{12} = 5K\)

\(\Rightarrow K = \frac{1}{60}\)

and,

\(\frac{70 - 60}{t} = K (\frac{70 + 60}{2} - 25)\)

\(\Rightarrow \frac{10}{t} = K (\frac{130}{2} - 25)\)   ----(2)

Now, on solving the equation (1) and (2) we have;

t = 15 min.

Hence, option 2) is the correct answer.