Solution 10
At maximum temperature (steady state), the rate of heat supplied by the heater ($P$) equals the rate of heat loss to the surroundings. $$ P = h_{coeff} A (\theta – \theta_0) $$ where $A$ is the total surface area and $\theta_0 = 20^\circ\text{C}$ is the ambient temperature.
Geometry Analysis:
The vessels are identical except for height. Since volumes are $V_1=1$, $V_2=2$, $V_3=3$, the heights must be in ratio $h:2h:3h$.
Total surface area for a closed cylinder (top, bottom, curved):
$$ A = 2\pi r^2 + 2\pi r h = 2\pi r (r + h) $$
Using the data for Vessel 1 ($80^\circ$C) and Vessel 2 ($60^\circ$C): $$ P = C \cdot A_1 (80 – 20) = 60 C A_1 $$ $$ P = C \cdot A_2 (60 – 20) = 40 C A_2 $$ Equating the power $P$: $$ 60 A_1 = 40 A_2 \implies 3 A_1 = 2 A_2 $$ Substituting the area formula (letting $h_1 = h$ and $h_2 = 2h$): $$ 3 [2\pi r(r+h)] = 2 [2\pi r(r+2h)] $$ $$ 3r + 3h = 2r + 4h \implies r = h $$ The radius of the vessel equals the height of the first vessel.
Calculating for Vessel 3:
Height $h_3 = 3h$. Area $A_3 = 2\pi h (h + 3h) = 8\pi h^2$.
Comparing to $A_1 = 2\pi h (h + h) = 4\pi h^2$, we find $A_3 = 2 A_1$.
Write the heat balance for Vessel 3 with unknown max temp $\theta_3$: $$ P = C A_3 (\theta_3 – 20) $$ Substitute $P = 60 C A_1$ and $A_3 = 2 A_1$: $$ 60 C A_1 = C (2 A_1) (\theta_3 – 20) $$ $$ 60 = 2 (\theta_3 – 20) $$ $$ 30 = \theta_3 – 20 \implies \theta_3 = 50^\circ\text{C} $$