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Hi Mark,

When you have water at two different temperatures and mix them the heat would flow from higher temperature (hot water) to the lower temperature. So the basic equation here is, heat lost by hot water = heat gained by cold water as both reach a thermal equilibrium.

Heat transfer **Q = mcΔT.**

Where **m** is the mass of the water, **c** is the specific heat (4.1855 J/g°C) and **ΔT** is the change in temperature.

For your case, X litres at 5 degrees C and Y litres at 10 degrees C,

Cold 5——x——10 Hot

(X 10^{3})(4.186)(10 – x) = (Y 10^{3})(4.186)(x – 5)

Solving this, we get

X/Y = x – 5 / 10 – x

Now for your other cases, where you need to find the ratio of hot and cold water to mix and attain a specific temperature (x), all you need to do is substitute the equilibrium temperature value for x. X/Y is the ratio of the water you need to mix.