The temperature of a body in air falls from 4 | vedclass

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(C) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{T_1 - T_2}{t} = K \left[ \frac{T_1 + T_2}{2} - T_s \right]$.Given: $T

Vedclass · Accepted Answer

$\frac{56}{3}^{\circ} 	ext{C}$

Vedclass · Answer

(C) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{T_1 - T_2}{t} = K \left[ \frac{T_1 + T_2}{2} - T_s ight]$.Given: $T_1 = 40^{\circ} 	ext{C}$,$T_2 = 24^{\circ} 	ext{C}$,$t = 4 	ext{ min}$,$T_s = 16^{\circ} 	ext{C}$.Substituting the values: $\frac{40 - 24}{4} = K \left[ \frac{40 + 24}{2} - 16 ight]$.$\frac{16}{4} = K [32 - 16] \implies 4 = K(16) \implies K = \frac{1}{4}$.Now,for the next $4 	ext{ minutes}$,let the final temperature be $T$. Here $T_1 = 24^{\circ} 	ext{C}$,$T_2 = T$,$t = 4 	ext{ min}$.$\frac{24 - T}{4} = \frac{1}{4} \left[ \frac{24 + T}{2} - 16 ight]$.$24 - T = \frac{24 + T - 32}{2} = \frac{T - 8}{2}$.$48 - 2T = T - 8 \implies 3T = 56 \implies T = \frac{56}{3}^{\circ} 	ext{C}$.

The temperature of a body in air falls from $40^{\circ} \text{C}$ to $24^{\circ} \text{C}$ in $4 \text{ minutes}$. The temperature of the air is $16^{\circ} \text{C}$. The temperature of the body in the next $4 \text{ minutes}$ will be $:$

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