A body takes 10 , text{minutes} to cool from | vedclass

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(A) According to $Newton's$ law of cooling, the rate of cooling is given by: $\frac{	heta_1 - 	heta_2}{t} = K \left( \frac{	heta_1 + 	heta_2}{2} -

Vedclass · Accepted Answer

$43$

Vedclass · Answer

(A) According to $Newton's$ law of cooling, the rate of cooling is given by: $\frac{	heta_1 - 	heta_2}{t} = K \left( \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight)$.For the first interval: $\frac{60 - 50}{10} = K \left( \frac{60 + 50}{2} - 25 ight) \implies 1 = K(55 - 25) \implies 1 = 30K \implies K = \frac{1}{30} \dots (i)$.For the second interval, let the final temperature be $	heta$: $\frac{50 - 	heta}{10} = K \left( \frac{50 + 	heta}{2} - 25 ight)$.Substituting $K = \frac{1}{30}$: $\frac{50 - 	heta}{10} = \frac{1}{30} \left( \frac{50 + 	heta - 50}{2} ight) = \frac{1}{30} \left( \frac{	heta}{2} ight) = \frac{	heta}{60}$.$6(50 - 	heta) = 	heta \implies 300 - 6	heta = 	heta \implies 7	heta = 300 \implies 	heta \approx 42.85 \, ^\circ	ext{C} \approx 43 \, ^\circ	ext{C}$.

$A$ body takes $10 \, \text{minutes}$ to cool from $60 \, ^\circ\text{C}$ to $50 \, ^\circ\text{C}$. The temperature of the surroundings is constant at $25 \, ^\circ\text{C}$. Then, the temperature of the body after the next $10 \, \text{minutes}$ will be approximately ....... $^\circ\text{C}$

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