Modeling with First Order DE’s — Question 2

Question 2 – Solution

Strategy. Model the difference between the object and its time-dependent surroundings; the ambient temperature is not a constant equilibrium value.

Step 1: Form and solve the IVP. Newton’s law gives $$T'=-0.10(T-20-2t),\qquad T(0)=80.$$ The right side is in degrees Celsius per minute. The object cools when it is hotter than the chamber and warms when it is colder. An integrating factor $e^{t/10}$ yields $$\boxed{T(t)=2t+80e^{-t/10}},\qquad t\ge 0.$$ Its derivative is $2-8e^{-t/10}$, which equals $-0.10(T-T_a)$, and its initial value is $80$.

Step 2: Identify the turning time. The derivative vanishes when $e^{-t/10}=1/4$, giving $$\boxed{t_m=10\ln 4\approx 13.86\text{ min}},\qquad \boxed{T_{\min}=20+20\ln 4\approx 47.73^\circ\mathrm C}.$$ Since $T''=0.8e^{-t/10}>0$, the derivative changes from negative to positive exactly once. At this time $T=T_a$; afterward the rising chamber is warmer than the object, so the heat-transfer term becomes positive.

Step 3: Interpret the long-time lag. The signed temperature difference is $$T-T_a=80e^{-t/10}-20\longrightarrow-20^\circ\mathrm C.$$ The object eventually warms at almost $2^\circ$C/min. A temperature deficit of $20^\circ$C supplies that rate because $-0.10(-20)=2$. Permanent equality would instead give zero heat-transfer rate, inconsistent with following an ambient temperature that keeps rising. The long-time conclusion is conditional on the stated linear ambient schedule and constant cooling coefficient remaining valid.