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Step-by-Step Solution
Step 1: State Newton’s Law of Cooling
Newton’s law of cooling states that the rate of change of temperature of a body is proportional to the difference between its own temperature and the ambient (surrounding) temperature. Mathematically, it can be written as
$ \frac{dT}{dt} = K \bigl(T - T_s\bigr), $
where:
$ T $ is the temperature of the body at time $ t $,
$ T_s $ is the surrounding (room) temperature,
$ K $ is a constant of proportionality.
Step 2: Identify the Important Temperature Points
The body cools from $ 3T $ to $ 2T $ in 10 minutes, where the surrounding temperature is $ T $. We want to find its temperature $ T' $ after the next 10 minutes.
Step 3: Express First Cooling Interval (3T to 2T)
During the first 10 minutes, the temperature drops from $ 3T $ to $ 2T $, so the average temperature of the body in this interval can be taken as
$ T_1 = \frac{3T + 2T}{2} = 2.5T. $
The rate of cooling over this interval is
$ \frac{\Delta T}{\Delta t} = \frac{3T - 2T}{10} = \frac{T}{10}. $
By Newton’s law of cooling for the first interval,
$ \frac{dT}{dt} = K \bigl(T_1 - T_s\bigr). $
Substitute $ T_1 = 2.5T $ and $ T_s = T $ to get
$ \frac{T}{10} = K(2.5T - T) = K(1.5T). $
Hence,
$ K = \frac{T/10}{1.5T} = \frac{1}{15}. $
Step 4: Express Second Cooling Interval (2T to T') in 10 Minutes
Let $ T' $ be the new temperature of the body at the end of the next 10 minutes. The average temperature of the body in this interval is
$ T_2 = \frac{2T + T'}{2}. $
The temperature drop in this interval is $ 2T - T' $ over 10 minutes, so
$ \frac{\Delta T}{\Delta t} = \frac{2T - T'}{10}. $
By Newton’s law of cooling for the second interval,
$ \frac{2T - T'}{10} = K \Bigl(\frac{2T + T'}{2} - T\Bigr). $
Step 5: Substitute the Value of K and Simplify
From Step 3, we have $ K = \frac{1}{15} $. Substitute this into the second interval’s equation:
$ \frac{2T - T'}{10} = \frac{1}{15} \left(\frac{2T + T'}{2} - T\right).
$
Multiply throughout by common denominators to simplify:
$ 15(2T - T') = 10 \left(\frac{2T + T'}{2} - T\right).
$
$ 15(2T - T') = 10 \left(\frac{2T + T' - 2T}{2}\right) = 10 \left(\frac{T'}{2}\right) = 5T'.
$ 30T - 15T' = 5T'
$
$ 30T = 20T'
$ T' = \frac{30T}{20} = \frac{3}{2} T.
Step 6: Conclude the Temperature After Next 10 Minutes
Thus, after the next 10 minutes, the temperature of the body will be
$ \frac{3}{2} T. $
This matches the given correct answer.
