Hot into cold.

Will this help heat transfer

It may, CJ. Thanks, I'll try to digest it.

It's pretty simple algebra if there is only one substance and there's no phase shift (is that the right term?). If the substances are different, though, or if it involves a change from liquid to gas or such, it gets a lot more complicated, and you have to (ugh) look things up.

The key is that the heat lost by the warmer water is equal to the heat gained by the cooler water. Taking some quick shortcuts (i.e. realizing that gallons are proportional to mass in this case and that specific heat capacity cancels out)...

The heat lost by the warmer water (which is mass * C * change_in temperature) is

50 gallons * C * (140 - T)

(The shortcut is OK because C is in Joules/(Gallon degree Fahrenheit).

The heat gained by the cooler water is

300 gallons * C * (T - 90)

So 50 * C (140-T) = 300 * C * (T - 90)

So the C's cancel out (thank goodness)... and the equation is

7000 - 50T = 300T - 27000

350T = 34000

T = 97.14 F

Who needs all that math? If it's too hot put an ice cube in it.