Correct option is (2) 13/5 t
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, T1 = 90°C
Final temperature, T2 = 80°C
and surrounding temperature, Ts = 20°C
Now, on putting the above values in equation (1) we get;
\(\frac{90 - 80}t= k \left(\frac{90 + 80}2 - 20\right)\)
⇒ \(k[85 - 20] = \frac{90 - 80}t\)
⇒ \(k[65] = \frac{10}t\)
⇒ \(k = \frac 2{13 t}\) ......(2)
For the second case we have;
Similarly,
Given: Initial Temperature, T1 = 80°C
Final temperature, T2 = 60°C
and surrounding temperature, Ts = 20°C
\(\frac{80 - 60}{t'} = k\left(\frac{80 + 60}2 - 20\right)\)
⇒ \(k[50] = \frac{20}{t'}\) ......(3)
where t' is the time taken in the second case
Substituting the equation (2) in (3) we get;
\(\frac 2{13t} (50) = \frac{20}{t'} \)
⇒ \(t' = \frac{13 t}5\)