Question

If heat goes into a liquid, do I need to care about the work done on the liquid? Aren't they incompressible?

Answer 1

Sure. It's small, but when you only have solids and liquids, it is not negligible in the big picture.

(When you have a gas involved though, the work done is considerably more significant than the work done on liquids and solids at the same applied external pressure.)

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Consider `"1 mol"` of liquid water in a closed system where we've reversibly heated it at `"1 atm"` of atmospheric pressure from `25^@ "C"` to `30^@ "C"`.

Although the heat flow `q` is nonneglible (around `"376 J"`), let's consider the work done. The molar volume of water at `25^@ "C"` is:

`"L"/"997.0479 g" xx "18.015 g"/"1 mol water"`

`=` `"0.018068 L/mol"`

A realistic new molar volume for the water after expansion and heating to `30^@ "C"` would be `"0.018095 L/mol"`. It would mean that the expansion work required was:

`color(blue)(w) = -PDeltaV = -P(V_2 - V_1)`

`= -("1 atm")("0.018095 L"/"mol" - "0.018068 L"/"mol")("1 mol water") xx "8.314472 J"/("0.082057 L"cdot"atm")`

`= color(blue)(-"0.002736 J")` of work.

(And back-calculating shows the new density would be `"995.5789 g/L"`, which is approximately the water density near `30^@ "C"`.)

That is evidently an extremely small amount of work for a `5^@ "C"` temperature change, compared to an ideal gas.

For instance, `"1 mol"` of an ideal gas being expanded from `25^@ "C"` to `30^@ "C"` would have the following change in volume at `"1 atm"` atmospheric pressure:

`V_1 = (nRT_1)/P = (("1 mol")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("1 atm")`

`=` `"24.4653 L"`

`V_2 = (nRT_2)/P = (("1 mol")("0.082057 L"cdot"atm/mol"cdot"K")("303.15 K"))/("1 atm")`

`=` `"24.8756 L"`

`color(green)(DeltaV_"gas") = V_2 - V_1 = color(green)("0.4103 L")`

compared to liquid water, which had `color(green)(DeltaV_"liq" = "0.000027 L")`.

The change in volume was over 15000 times as significant for a gas as for a liquid.