Heating water

Consider a small conical strip in the material of the container of width $dx$ such that the temperature difference between its ends be $d\theta$ .

Applying the formula, $\dfrac{Q}{t} = \dfrac{kA(T_1 - T_2)}{l}$ .

$\begin{aligned} \dfrac{Q}{t} & = \dfrac{k(\pi x \sqrt{x^2 + h^2}) d\theta}{dx}\\ \\ \dfrac{1}{x\sqrt{h^2 + x^2}} dx & = \dfrac{k\pi t}{Q} d\theta\\ \\ \int_{r_1}^{r_2} \dfrac{1}{x\sqrt{h^2 + x^2}} dx & = \int_{T_1}^{T_2} \dfrac{k\pi t}{Q} d\theta\\ \\ {\left[\dfrac{\ln(x) - \ln\left(h\left(\sqrt{h^2 + x^2} + h\right)\right)}{h}\right]}_{r_1}^{r_2} & = {\left[\dfrac{k\pi t \theta}{Q}\right]}_{T_1}^{T_2}\\ \\ \dfrac{\ln\left(\dfrac{r_2\left(\sqrt{h^2 + {r_1}^2} + h\right)}{r_1\left(\sqrt{h^2 + {r_2}^2} + h\right)}\right)}{h} & = \dfrac{k\pi t\left(T_2 - T_1\right)}{Q}\\ \\ \dfrac{Q}{t} & = \dfrac{k\pi h\left(T_2 - T_1\right)}{\ln\left(\dfrac{r_2\left(\sqrt{h^2 + {r_1}^2} + h\right)}{r_1\left(\sqrt{h^2 + {r_2}^2} + h\right)}\right)} \end{aligned}$

Substituting the values, we get:-
$\begin{aligned} \dfrac{Q}{t} & = \dfrac{0.5 × \pi × 1 × (80 - 10)}{\ln\left(\dfrac{3\left(\sqrt{1^2 + 2^2} + 1\right)}{2\left(\sqrt{1^2 + 3^2} + 1\right)}\right)}\\ \\ & = \color{#3D99F6}{\boxed{715.105}} \end{aligned}$

The answer I am getting is 709.42 . The question accepted this value . If you all agree with my answer , I shall publish a report .

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Taking a frustrum of bottom radius r , you will get the surface are through which heat is emitted as

$A = 2 \pi (1 - r) \sqrt {5}$

The process to calculate this goes as -

Imagine complete cone , with upper imaginary conical part to complete cone .

Let the angle which the botom makes with cone's inclination be $\theta$ .

$tan\theta = 0.5$

$S = \frac {\pi r^2 }{cos\theta} - \frac {\pi (r - 2)^2}{cos\theta}$

Radius of upper imaginary part of cone can be calculated easily by trigonometry.

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Now using

$Power =\frac {K A \Delta T }{dr }$

Answer comes as $P = \frac {70 \pi \sqrt {5}}{ln 2}$