A body cools from 62^{circ}C to 50^{circ}C in | vedclass

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

Vedclass · Accepted Answer

$26$

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(B) According to Newton's law of cooling,the rate of cooling is proportional to the difference in temperature between the body and its surroundings:$\frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_s ight)$For the first interval ($62^{\circ}C$ to $50^{\circ}C$ in $10 	ext{ minutes}$):$\frac{62 - 50}{10} = k \left( \frac{62 + 50}{2} - T_s ight)$$1.2 = k (56 - T_s) \quad \dots (1)$For the second interval ($50^{\circ}C$ to $42^{\circ}C$ in $10 	ext{ minutes}$):$\frac{50 - 42}{10} = k \left( \frac{50 + 42}{2} - T_s ight)$$0.8 = k (46 - T_s) \quad \dots (2)$Dividing equation $(1)$ by equation $(2)$:$\frac{1.2}{0.8} = \frac{56 - T_s}{46 - T_s}$$1.5 = \frac{56 - T_s}{46 - T_s}$$1.5(46 - T_s) = 56 - T_s$$69 - 1.5T_s = 56 - T_s$$69 - 56 = 1.5T_s - T_s$$13 = 0.5T_s$$T_s = 26^{\circ}C$Thus,the temperature of the surroundings is $26^{\circ}C$.

$A$ body cools from $62^{\circ}C$ to $50^{\circ}C$ in $10 \text{ minutes}$ and to $42^{\circ}C$ in the next $10 \text{ minutes}$. The temperature of the surroundings is ........ $^{\circ}C$.

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