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If surrounding air is kept at $20^{\circ} \mathrm{C}$ and body cools from $80^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in 5 minutes, then the temperature of the body after 15 minute will be
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The correct answer is:
$54.7^{\circ} \mathrm{C}$
$\frac{d T}{d t}=-k(T-20) \Rightarrow T-20=e^{-k t+c} \Rightarrow T=20+e^c e^{-k t}$
for $t=0, T=80 \Rightarrow e^c=60$
i.e. $T=20+60 \cdot e^{-k t}$
For $t=5, T=70 \Rightarrow 70=20+60 e \Rightarrow-5 k=\log \frac{5}{6}$
i.e. $T=20+60 e^{\frac{1}{5} \log \frac{5}{6} \times t}$
Now, for $t=15$
$T=20+60 e^{15 \times \frac{1}{5} \log \frac{5}{6}}=20+60 \times\left(\frac{5}{6}\right)^3=54.7$
for $t=0, T=80 \Rightarrow e^c=60$
i.e. $T=20+60 \cdot e^{-k t}$
For $t=5, T=70 \Rightarrow 70=20+60 e \Rightarrow-5 k=\log \frac{5}{6}$
i.e. $T=20+60 e^{\frac{1}{5} \log \frac{5}{6} \times t}$
Now, for $t=15$
$T=20+60 e^{15 \times \frac{1}{5} \log \frac{5}{6}}=20+60 \times\left(\frac{5}{6}\right)^3=54.7$
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