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A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be:
A
$\frac{4}{3}T$
B
T
C
$\frac{7}{4}T$
D
$\frac{3}{2}T$
Explanation
Use the average-temperature form of Newton's law twice.
Detailed Solution
Newton's law of cooling: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T\right)$
$\frac{3T - 2T}{10} = k\left(\frac{5T}{2} - T\right) \Rightarrow \frac{T}{10} = k\left(\frac{3T}{2}\right)$ ...(i)
$\frac{2T - T'}{10} = k\left(\frac{2T + T'}{2} - T\right) = k\left(\frac{T'}{2}\right)$ ...(ii)
Solving (i) and (ii): $T' = \frac{3}{2}T$
$\frac{3T - 2T}{10} = k\left(\frac{5T}{2} - T\right) \Rightarrow \frac{T}{10} = k\left(\frac{3T}{2}\right)$ ...(i)
$\frac{2T - T'}{10} = k\left(\frac{2T + T'}{2} - T\right) = k\left(\frac{T'}{2}\right)$ ...(ii)
Solving (i) and (ii): $T' = \frac{3}{2}T$