Let's say I have ground water at a depth of $3000m$ at $50^{o}C$.

If I now pump this water up to the surface, the water pressure decreases from the $\pm 300 atm$ at that depth to $1 atm$ at the surface.

Silly enough I thought of using the ideal gas law, but as it's name implies it's only valid for ideal gasses, of which a basically incompressible liquid is not one.

If I use the ideal gas law for a isovolumetric process where V is constant, I get:

$T_1=T_2\frac{P_1}{P_2}$

which yields a rediculously small value for $T_1$ which cannot be right.

How would I go about calculating the change in water temp as a function of depth?

I am sure it won't be massive, but I think in the calculation I need it for, where I have a heat engine that already has a Carnot efficiency of $9\%$ if I assume $\Delta T$ is negligible I think that because of the great depth involved the temperature might lower sufficiently as it is pumped up that this efficiency might be even lower.

## Best Answer

As you calculated, but seem to dismiss, the temperature change in water pumped up from deep, high-pressure zones to the surface depends very little on the change in pressure. Other factors would have far more effect:

If you're thinking to run a heat engine from the difference between the raised water and the local environment, you'll need to take these factors into account when calculating the efficiency.