Modeling — Question 3

Question 3 – Solution

Strategy. Write heat-capacity balances before shifting the reference temperature; sum and difference variables isolate two decay rates.

Step 1: Balance heat for each body. For body 1, capacity times $T_1'$ equals $(T_2-T_1)+(20-T_1)$ in kJ/min; the analogous balance holds for body 2. Hence $$\boxed{\theta_1'=-2\theta_1+\theta_2,\qquad \theta_2'=\theta_1-2\theta_2,\qquad (\theta_1(0),\theta_2(0))=(60,0).}$$ The numerical decay coefficients have units min$^{-1}$.

Step 2: Track total excess energy. With each capacity equal to $1$ kJ/$^\circ$C, excess energy is $E=\theta_1+\theta_2$ numerically in kJ. Interbody exchange cancels, leaving $E'=-E$ and $\boxed{E(t)=60e^{-t}\text{ kJ}}$. Only heat delivered to the room decreases this total.

Step 3: Solve the modes and locate the warm-up peak. The difference obeys $(\theta_1-\theta_2)'=-3(\theta_1-\theta_2)$. Together with the sum this gives $$\boxed{T_1=20+30(e^{-t}+e^{-3t}),\quad T_2=20+30(e^{-t}-e^{-3t}).}$$ Here $T_2'=30(-e^{-t}+3e^{-3t})$, which changes sign from positive to negative at $t_*=(\log 3)/2$. Therefore $$\boxed{T_{2,\max}=(20+20/\sqrt 3)\ ^\circ\mathrm C,\qquad t_*=(\log 3)/2\text{ min}.}$$

Step 4: Reconcile local warming with total cooling. Initially the hotter body supplies heat to body 2 faster than body 2 loses heat to the room. Nevertheless $E'<0$ throughout. The formulas give $\theta_1,\theta_2\ge 0$ and $\theta_1+\theta_2=60e^{-t}\le 60$, so each temperature stays in $[20,80]^\circ$C. Both tend to room temperature; the second has a genuine temporary maximum, not a higher final equilibrium.