# Boilling a liquid in a closed container, less heat required?

If you boil a liquid with latent heat $L$ in a container of infinite volume (i.e. open to the universe) the heat required is: $$Q=Lm$$ Where $m$ is the mass of the liquid boiled. If however you boil it in a container of finite volume, $V$, the heat required is: $$Q=Lm-V\Delta p$$ Where $\Delta p$ is the change in pressure. By what physical mechanism is the system given the heat $V \Delta p$ back to the surrounding and why?

Edit

Let me explain my reasoning behind the formula $Q=Lm-V\Delta p$. By definition the latent heat of vaporization is the difference in the enthalpy's between the evaporated gas and the liquid i.e. $$Lm=H_2-H_1$$ But $H_i=U_i+p_i V$ for $i=1,2$ therefore: $$Lm=(U_2+p_2V)-(U_1+p_1V)$$ $$Lm=\Delta U + V\Delta p$$ And thus: $$\Delta U=Lm-V\Delta p$$ In our case there is no work been done on the system from the surroundings and hence the change in internal energy must solely be provided by the addition of heat energy. This means that the heat energy required for this change is: $$Q=Lm-V\Delta p$$

Edit 2

I think there is a related concept here, that is worth noting. Take a system at $T_1, p_1, V_1$ and change it to the state $T_2, p_1, V_2$ via a reversible change. I am guessing we would all agree that the heat supplied to the system is given by: $$Q=C_p (T_2-T_1)$$ But the internal energy is given by: $$dU=dQ-pdV$$ So the change in internal energy is given by: $$U=C_p(T_2-T_1)-p_1(V_2-V_1)$$ Now again starting at $T_1, p_1, V_1$ and finishing at $T_2, p_1, V_2$ we instantaneously (e.g. by opening a tap) increase the volume from $V_1$ to $V_2$. Then act to change the temperature since no work is done on the system at any stage the heat applied must equal the change in internal energy of the system, i.e. $$Q_2= C_p(T_2-T_1)-p_1(V_2-V_1)$$ Of course this is exactly the same argument as above, but I think using heat capacities simplifies it intuitively.

• Where did you get that equation? I would think the - should be a +. – paparazzo Oct 29 '15 at 10:16
• @Frisbee $\Delta H=\Delta U+V \Delta P=Lm$ – Quantum spaghettification Oct 29 '15 at 13:36
• @Frisbee See physics.stackexchange.com/questions/215365/… – Quantum spaghettification Oct 29 '15 at 18:37
• The heat required is less because the volume of the container is fixed and the pressure increases. This increases the rate of collision between the gas molecules and water surface. This supplies the remaining energy for vaporisation. – ShankRam Nov 1 '15 at 7:56
• I don't think in the second case you can consider the change of enthalpy equal to the heat, since you are considering a finite volume and with the boiling process ongoing the pressure will unlikely remain constant – NNec Nov 1 '15 at 9:37

Imagine an adiabatic container (like a Dewar bottle): perfectly insulated and pressure resistant, filled with water. A heating element provides a heat flow into the container but no heat can escape from it. No work can be done by the system either: the volume is constant. See diagram: Now if we add some heat to the system by means of the heating element and at some arbitrary point stop adding heat, the system (free of losses and incapable of performing work) remains at constant $p,T,U$ values, basically forever. The OPs statement:

By what physical mechanism is the system given the heat VΔp back to the surrounding and why?

... therefore makes no sense to me.

By means of the Mollier diagram (pressure versus Enthalpy, see below) of water/steam let's look at what really happens when water is boiled in conditions of rising pressure. When we reach $0.1\:\mathrm{MPa}$, $100\:\mathrm{Celsius}$ at point red A ($0\:\mathrm{\%}$ dryness means all liquid and no vapour) and continue to add heat, we move to point green B ($100\:\mathrm{\%}$ dryness means no liquid, all vapour). From red A to green b we expend an amount of Enthalpy of $2247\:\mathrm{kJ/kg}$, that is the tabled Enthalpy of Vapourisation at $1\:\mathrm{atm}$, $100\:\mathrm{Celsius}$.

Now construct similar points A and B at higher temperature and corresponding pressure and note how the Enthalpy of Vapourisation gradually decreases, until in the Critical Point of $374\:\mathrm{Celsius}$ and $2.2\:\mathrm{MPa}$ it completely vanishes.

See also this diagram which shows the Enthalpy of Vapourisation for a few substances including water in function of temperature: Here too we can see that the Enthalpy of Vapourisation for water is far from constant and tends to zero towards the Critical Point.

First derivation: the equation $Lm=H_2-H_1$ is only valid for a process at p=constant, which is not your case.

Second derivation: the equation $Q=C_p (T_2-T_1)$ is only valid for a process at p=constant, which is not your case. Plus the rest of the argument doesn't make sense: both $\Delta T=0$ and $\Delta V=0$ in a closed container with boiling water.