Alan

If by "25 feet long" you are referring to the horizontal distance then the distance along the sloping surface can be found from Pythagoras:

$${\frac{{\sqrt{{{\mathtt{25}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{12}}\right)}^{{\mathtt{2}}}}}}{{\mathtt{12}}}} = {\mathtt{25.086\: \!655\: \!372\: \!483\: \!948\: \!9}}$$  feet

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100% of 20 = 20

10% of 20 = 2

40% of 20 = 4*10% of 20 = 4*2 = 8

2.5% of 20 = 1/4 * 10% of 20 = 1/4 * 2 = 0.5

Add up the numbers in bold to get 142.5% of 20 = 20+8+0.5 = 28.5

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I don't know that mathematics has a father; but it has a prince!  

Here's a numerical approach:

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The energy required is given by mass*specific heat capacity*temperature rise

The specific heat capacity of water (in this temperature range) is 4184 J/(kg.degC)

To answer your question you need to specify the starting temperature of the water as well, as it is the temperature difference that matters.

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If 2^x = 7 then x = log₂(7)

$${\mathtt{x}} = {{log}}_{{\mathtt{2}}}{\left({\mathtt{7}}\right)} \Rightarrow {\mathtt{x}} = {\mathtt{2.807\: \!354\: \!922\: \!057\: \!604}}$$

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Volume = mass/density

Volume = 15000 g/45 g/cm³ ≈ 333.3 cm³

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Here's a simple example comparison.

Non-implicit ( y = f(x) )

y = x + 1          

dy/dx = 1

Implicit ( y = f(x,y) )

y = x*y + 1

dy/dx = y*1 + x*dy/dx

(1 - x)dy/dx = y

dy/dx = y/(1 - x)

But (1 - x)y = 1, or y = 1/(1 - x), so, substituting this in the above, we get dy/dx = 1/(1-x)²

In this case, from y = 1/(1-x) we could have written directly;  dy/dx = -(1-x)^-2*(-1)  or  dy/dx = 1/(1-x)², the same as above.  

However, we can't always rearrange y = f(x,y) to get y = f(x) explicitly, so sometimes we are forced to use implicit differentiation.

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Assuming the diameter is 2ft and the height is 4ft the volume is pi*1²*4 ft³ or pi*4*12³ in³ 

Since the US is about the only place left using gallons regularly, we'll take the US gallon, which is defined as 231 in³, so the volume is 

$${\mathtt{volume}} = {\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{12}}}^{{\mathtt{3}}}}{{\mathtt{231}}}} \Rightarrow {\mathtt{volume}} = {\mathtt{94.002\: \!980\: \!180\: \!141\: \!345\: \!7}}$$

or volume ≈ 94 gallons

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