A spherical metal ball at 80^{circ} C cools i | vedclass

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(D) Let $	heta$ be the temperature of the ball at any time $t$. According to Newton's law of cooling,$\frac{d	heta}{dt} = -k(	heta - 20)$,where $k

Vedclass · Accepted Answer

$31.85$

Vedclass · Answer

(D) Let $	heta$ be the temperature of the ball at any time $t$. According to Newton's law of cooling,$\frac{d	heta}{dt} = -k(	heta - 20)$,where $k > 0$.Integrating both sides,we get $\ln|	heta - 20| = -kt + C$.At $t = 0$,$	heta = 80^{\circ} C$,so $C = \ln(80 - 20) = \ln(60)$.Thus,$\ln|	heta - 20| = -kt + \ln(60) \dots (i)$.At $t = 5$,$	heta = 60^{\circ} C$,so $\ln(60 - 20) = -5k + \ln(60)$.$5k = \ln(60) - \ln(40) = \ln(\frac{60}{40}) = \ln(\frac{3}{2})$.So,$k = \frac{1}{5} \ln(\frac{3}{2})$.For $t = 20$,substituting $k$ into equation $(i)$:$\ln|	heta - 20| = -20 	imes \frac{1}{5} \ln(\frac{3}{2}) + \ln(60) = -4 \ln(\frac{3}{2}) + \ln(60)$.$\ln|	heta - 20| = \ln((\frac{2}{3})^4) + \ln(60) = \ln(\frac{16}{81} 	imes 60) = \ln(\frac{16 	imes 20}{27}) = \ln(\frac{320}{27}) \approx \ln(11.85)$.$	heta - 20 = 11.85 \implies 	heta = 31.85^{\circ} C$.

$A$ spherical metal ball at $80^{\circ} C$ cools in $5 \text{ minutes}$ to $60^{\circ} C$ in a surrounding temperature of $20^{\circ} C$. The temperature of the ball after $20 \text{ minutes}$ is approximately: (in $^{\circ} C$)

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