Quarter-load heat run and 20% overload time

For heating from cold, the temperature rise after time $t$ is $\theta = \theta_m\left(1-e^{-t/\tau}\right)$, where $\theta_m$ is the final steady rise and $\tau$ the heating time constant. The final rise is proportional to the total losses ($\theta_m \propto P_i + x^2P_c$).

Step 1: Time constant and final rise at quarter load

Let $x = e^{-1/\tau}$ ($\tau$ in hours). Then

$$\begin{aligned} 40 &= \theta_{mq}(1-x^3), \qquad 50 = \theta_{mq}(1-x^4)\\ \frac{50}{40} &= \frac{1-x^4}{1-x^3} \;\Rightarrow\; x^4 - 1.25x^3 + 0.25 = 0\\ (x-1)(x^3 - 0.25x^2 - 0.25x - 0.25) &= 0 \end{aligned}$$

$x = 1$ is not acceptable, so solving the cubic: $x = 0.8689$.

$$\begin{aligned} \tau &= \frac{1}{\ln(1/0.8689)} = 7.11\ \text{h}\\ \theta_{mq} &= \frac{40}{1-0.8689^3} = \frac{40}{0.344} = 116.26\ ^\circ\text{C} \end{aligned}$$

Check: $116.26(1-0.8689^4) = 50\ ^\circ$C.

Step 2: Losses at different loads

Maximum efficiency at 65% load means iron loss = copper loss at that load:

$$P_i = (0.65)^2P_c = 0.4225\,P_c$$

Step 3: Final steady rise at full load

$$\theta_{mf} = 116.26 \times \frac{1.4225}{0.4850} = 341.0\ ^\circ\text{C}$$

Step 4: Time on 20% overload to reach the same rise

Final rise on overload:

$$\theta_{mo} = 116.26 \times \frac{1.8625}{0.4850} = 446.5\ ^\circ\text{C}$$

Time to reach 341.0 °C (the full-load final rise) on overload, starting cold:

$$\begin{aligned} 341.0 &= 446.5\left(1-e^{-t/7.11}\right)\\ e^{-t/7.11} &= 1-\frac{341.0}{446.5} = 0.2363\\ t &= 7.11\ln\frac{1}{0.2363} = 10.27\ \text{h} \end{aligned}$$

Answer: final steady rise at full load = 341.0 °C; time on 20% overload ≈ 10.27 h (about 616 min); $\tau$ = 7.11 h.

(The given data lead to a very high rise and long time constant; the method is what matters.)