A spherical metal ball at $80^{\circ} \mathrm{C}$ cools in 5 minutes to $60^{\circ} \mathrm{C}$, in surrounding temperature of $20^{\circ} \mathrm{C}$, then the temperature of the ball after 20 minutes is approximately

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Aicharya · Accepted Answer

Let $	heta$ be the temperature of ball at any time ' $t$ '. $\begin{aligned} & 	herefore \quad \frac{\mathrm{d} 	heta}{\mathrm{dt}} \propto(	heta-20) \ & \quad \Rightarrow \frac{\mathrm{d} 	heta}{\mathrm{dt}}=-\mathrm{k}(	heta-20), \mathrm{k}\gt0 \end{aligned}$ Integrating on both sides, we get $\log |	heta-20|=-\mathrm{kt}+\mathrm{c}$ when $\mathrm{t}=0, 	heta=80^{\circ}$ $\begin{array}{ll} 	herefore & c=\log 60 \ 	herefore & \log |	heta-20|=-k t+\log 60...(i) \end{array}$ $\begin{aligned} & \quad 	ext { When } \mathrm{t}=5, 	heta=60^{\circ} \ & 	herefore \quad \log 40=-5 \mathrm{k}+\log 60 \ & \Rightarrow 5 \mathrm{k}=\log 60-\log 40 \ & \Rightarrow 5 \mathrm{k}=\log \left(\frac{3}{2}ight) \ & \Rightarrow \mathrm{k}=\frac{1}{5} \log \left(\frac{3}{2}ight)\end{aligned}$ $	herefore \quad \log |	heta-20|=\frac{-1}{5} \log \left(\frac{3}{2}ight) \mathrm{t}+\log 60$ ...[From (i)] when $\mathrm{t}=20$, $\begin{aligned} & \Rightarrow \log |	heta-20|=\frac{-1}{5} \log \left(\frac{3}{2}ight) 20+\log 60 \ & \Rightarrow \log |	heta-20|=-4 \log \left(\frac{3}{2}ight)+\log 60 \ & \Rightarrow \log |	heta-20|=\log \left(\frac{2}{3}ight)^4+\log 60 \ & \Rightarrow \log |	heta-20|=\log \left(\frac{16 	imes 60}{81}ight) \ & \Rightarrow \log |	heta-20|=\log (11.85) \ & \Rightarrow 	heta-20=11.85 \ & \Rightarrow 	heta=11.85+20 \ & \Rightarrow 	heta=31.85 \end{aligned}$ The temperature of ball after 20 minutes is $31.85^{\circ} \mathrm{C}$

A spherical metal ball at $80^{\circ} \mathrm{C}$ cools in 5 minutes to $60^{\circ} \mathrm{C}$, in…

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