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Question: It takes 10 minutes to cool a liquid from 61\(^\circ\)C to 59\(^\circ\)C. If room temperature is 30\...

It takes 10 minutes to cool a liquid from 61∘^\circC to 59∘^\circC. If room temperature is 30∘^\circC then time taken in cooling from 51∘^\circC to 49∘^\circC is
(A) 10 min
(B) 11 min
(C) 13 min
(D) 15 min

Explanation

Solution

When a liquid cools down in a surrounding, the rate of cooling of the liquid through radiation is directly proportional to the difference in temperature of the liquid and the surrounding. Differential equations are generally used to describe this effect.
Formula used:
According to the Newton's law of cooling:
dTdt=C(T−T0)\dfrac{dT}{dt} = C(T - T_0)
where T0T_0 is the temperature of the surroundings.
The same expression in exponentials is:
T−T0=exp⁡−ktT - T_0 = \exp{-kt}

Complete answer:
The differential equation for Newton's law can also be written in the form of exponentials obtained after solving the differential equation as:
T2−T0T1−T0=exp⁡−k(t1−t2)\dfrac{T_2 - T_0}{T_1 - T_0} = \exp{-k ( t_1 - t_2 ) }
Now, the given temperatures for cooling in the first case are:
T2=61∘T_2 = 61 ^\circ C.
T1=59∘T_1 = 59 ^\circ C.
T0=30∘T_0 = 30 ^\circ C.
Therefore, the difference in the temperature becomes:
T2−T0=31∘T_2 - T_0 = 31^\circ C.
T1−T0=29∘T_1 - T_0 = 29^\circ C.
The time difference is given as:
t2−t1=10t_2 - t_1 = 10 min.
Substituting these values in the previous equation to obtain the value of k,
ln⁡3129=−k(10)\ln{ \dfrac{31}{29}} = -k (10)
  ⟹  −k=110ln⁡3129\implies -k = \dfrac{1}{10} \ln{ \dfrac{31}{29}}
Similarly, the time it takes for cooling from 51∘^\circ C to 49∘^\circ C is expressed as:
ln⁡2119=−kΔt\ln{ \dfrac{21}{19} }= -k \Delta t
On substituting the value of -k here and transposing , we get:
Δt=ln⁡2119110ln⁡3129=10×1.5006\Delta t = \dfrac{\ln{ \dfrac{21}{19} }}{\dfrac{1}{10} \ln{ \dfrac{31}{29}}} = 10 \times 1.5006 = 15 minutes nearly.

The correct answer is option (D).

Note:
Changing the differential equation to an expression containing exponentials involves integrals. Upon integration, a constant of integration called k appears in the expression. We have taken the expressions involving two temperatures separately and then divided the two expressions to obtain an expression that could help in eliminating the common constant. The point to remember here is that the expression involves difference of the temperature of the liquid and the surroundings.