A body cools from a temperature 3T to 2T in 1 | vedclass

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(D) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and the surroundings:$\fra

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$\frac{3}{2}T$

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(D) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and the surroundings:$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$For the first interval of $10$ minutes:$T_1 = 3T, T_2 = 2T, T_s = T, t = 10$$\frac{3T - 2T}{10} = K \left( \frac{3T + 2T}{2} - T \right)$$\frac{T}{10} = K \left( 2.5T - T \right) = K(1.5T) \implies K = \frac{1}{15} \dots (i)$For the next interval of $10$ minutes:Let the final temperature be $T'$.$T_1 = 2T, T_2 = T', T_s = T, t = 10$$\frac{2T - T'}{10} = K \left( \frac{2T + T'}{2} - T \right)$$\frac{2T - T'}{10} = K \left( \frac{T'}{2} \right) \dots (ii)$Substituting $K = \frac{1}{15}$ into equation $(ii)$:$\frac{2T - T'}{10} = \frac{1}{15} \left( \frac{T'}{2} \right)$$\frac{2T - T'}{10} = \frac{T'}{30}$$3(2T - T') = T'$$6T - 3T' = T'$$6T = 4T' \implies T' = \frac{6}{4}T = \frac{3}{2}T$

$A$ 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 the next $10$ minutes will be

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