A body initially at 80^circ C cools to 64^cir | vedclass

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(A) According to Newton's law of cooling,$\frac{	heta_1 - 	heta_2}{t} = K \left[ \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight]$.For the first int

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$42.7$

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(A) According to Newton's law of cooling,$\frac{	heta_1 - 	heta_2}{t} = K \left[ \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight]$.For the first interval ($0$ to $5$ min): $\frac{80 - 64}{5} = K \left[ \frac{80 + 64}{2} - 	heta_0 ight] \implies 3.2 = K(72 - 	heta_0)$ ... $(i)$For the second interval ($0$ to $10$ min): $\frac{80 - 52}{10} = K \left[ \frac{80 + 52}{2} - 	heta_0 ight] \implies 2.8 = K(66 - 	heta_0)$ ... $(ii)$Dividing $(i)$ by $(ii)$: $\frac{3.2}{2.8} = \frac{72 - 	heta_0}{66 - 	heta_0} \implies \frac{8}{7} = \frac{72 - 	heta_0}{66 - 	heta_0}$.$8(66 - 	heta_0) = 7(72 - 	heta_0) \implies 528 - 8	heta_0 = 504 - 7	heta_0 \implies 	heta_0 = 24^\circ C$.Substituting $	heta_0 = 24$ in $(i)$: $3.2 = K(72 - 24) \implies 3.2 = K(48) \implies K = \frac{3.2}{48} = \frac{1}{15}$.For the third interval ($0$ to $15$ min): $\frac{80 - 	heta}{15} = K \left[ \frac{80 + 	heta}{2} - 24 ight]$.$\frac{80 - 	heta}{15} = \frac{1}{15} \left[ \frac{80 + 	heta - 48}{2} ight] \implies 80 - 	heta = \frac{32 + 	heta}{2}$.$160 - 2	heta = 32 + 	heta \implies 3	heta = 128 \implies 	heta = 42.66^\circ C \approx 42.7^\circ C$.

$A$ body initially at $80^\circ C$ cools to $64^\circ C$ in $5$ minutes and to $52^\circ C$ in $10$ minutes. The temperature of the body after $15$ minutes will be ...... $^\circ C$.

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