Differential Equations
Newton's law of cooling
Grade Class 12
Question:
<p>Body cools 80°C→60°C in 10 min, room 25°C. Time to cool 60°C→40°C?</p>
<span>\(10\ln\frac{35}{7}\,\text{min}\)</span>
<span>\(10\frac{\ln(7/3)}{\ln(11/7)}\,\text{min}\)</span>
<span>\(5\ln(7/3)\,\text{min}\)</span>
<span>\(10\ln\frac{7}{3}\,\text{min}\)</span>
Step-by-Step Solution
Key Concept: Newton's law: dT/dt = -k(T-25). Solve and find k from first condition.
<div class='solution'><p>$T-25=Ae^{-kt}$. At $t=0$: $A=55$. $T=25+55e^{-kt}$. At $t=10$: $60=25+55e^{-10k}$ → $35=55e^{-10k}$ → $e^{-10k}=7/11$ → $k=-\ln(7/11)/10=\ln(11/7)/10$. For $60\to40$: $40=25+35e^{-k(t_1+10)}$... Starting from 60 at $t=10$: $35=55e^{-10k}$. From 60: $T_1-25=35$, to 40: $T_2-25=15$. $35e^{-k\tau}=15$ → $\tau=\ln(7/3)/k=10\ln(7/3)/\ln(11/7)$. <strong>Answer: (2)</strong>.</p></div>
Correct Answer: 2