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.)

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.

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Answer 2

Even though liquids are generally considered to be incompressible, you still need to consider the work done on the liquid when heat is added. This is because liquids can still change volume slightly in response to changes in pressure or temperature, albeit to a much smaller extent compared to gases. Additionally, the work done on a liquid may contribute to changes in its internal energy, affecting its temperature and other properties. Therefore, it's important to take into account both the heat added to the liquid and any work done on it to fully understand the changes occurring within the system.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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