Water at 100^{circ} C cools in 10 text{ minut | vedclass

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(A) According to Newton's Law of Cooling,$\frac{dT}{dt} = -k(T - T_s)$,where $T_s = 25^{\circ} C$. Integrating this,we get $T(t) = T_s + (T_0 - T_s)e^

Vedclass · Accepted Answer

$65.33$

Vedclass · Answer

(A) According to Newton's Law of Cooling,$\frac{dT}{dt} = -k(T - T_s)$,where $T_s = 25^{\circ} C$. Integrating this,we get $T(t) = T_s + (T_0 - T_s)e^{-kt}$. Given $T_0 = 100^{\circ} C$,so $T(t) = 25 + 75e^{-kt}$. At $t = 10 	ext{ min}$,$T = 80^{\circ} C$: $80 = 25 + 75e^{-10k} \Rightarrow 55 = 75e^{-10k} \Rightarrow e^{-10k} = \frac{55}{75} = \frac{11}{15}$. At $t = 20 	ext{ min}$,$T = 25 + 75e^{-20k} = 25 + 75(e^{-10k})^2$. Substituting the value: $T = 25 + 75 	imes (\frac{11}{15})^2 = 25 + 75 	imes \frac{121}{225} = 25 + \frac{121}{3} = 25 + 40.33 = 65.33^{\circ} C$.

Water at $100^{\circ} C$ cools in $10 \text{ minutes}$ to $80^{\circ} C$ in a room temperature of $25^{\circ} C$. What will be the temperature of water after $20 \text{ minutes}$ (in $^{\circ} C$)?

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