If I have 1 cup of water on the stove and another cup of water with a teaspoon of salt.
would the salt change the boiling time of the water?
[Physics] Does salt affect the boiling time of water
phase-transitionphysical-chemistrytemperaturethermodynamicswater
Related Solutions
Okay, first we have the phenomenon: Yes. adding salt increases the boiling point of water, which means that you have to input more energy to get the water to boil, but your egg or pasta will cook faster once you do, because the water will be hotter.
Then there's the why. The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is the same as the atmospheric pressure above the liquid. If we can artificially increase the vapor pressure of the liquid, we decrease the boiling temperature. If we can artificially decrease the vapor pressure of the liquid, we increase the boiling temperature. So the question has now become: why does the vapor pressure of water decrease when we add salt to it?
So imagine a pot of water. At any given temperature there will be some water molecules in the gas phase above the pot (that's the origin of the vapor pressure), and some in the liquid phase in the pot. The proportion in the two phases is determined by the interplay of lowering potential energy (by decreasing elevation in gravity, by forming hydrogen bonds, by lining up the polar ends of the molecules, etc.) and increasing the entropy (there's more accessible states in the gas phase, most liquids are incompressible, etc.). The potential energy part favors the liquid phase, while the entropy part favors the gas phase. The real requirement here is to minimize the free energy, F = U - TS, with F the free energy, U the potential, T the temperature, and S the entropy. Since S is paired with the temperature, increasing the temperature increases the impact of the entropy part, which is why the vapor pressure increases as we increase the temperature.
So now we toss in some salt, while keeping the temperature fixed. The volume fraction of the water decreases, and suddenly there are new accessible states for the water molecules in the liquid phase -- so the vapor pressure decreases. We keep adding salt and the vapor pressure keeps decreasing. If we keep going, eventually there's no vapor pressure.
Raoult's law says that the vapor pressure of a solution is proportional to the vapor pressure of the pure solvent (basically that there is a straight line between the pure vapor pressure and zero, when we've buried it in salt). That's taken as the definition of an ideal solution. Real solutions have a curved functional form between the two boundary conditions, with the deviations from linearity coming from interactions between the solute (the salt) and the solvent (the water). Those interactions might be things like breaking up the network of hydrogen bonds in the water, disrupting the polarization arrangement (both of which will favor gas phase), or bonding/pairing up with water molecules (which will favor liquid phase). At relatively low concentrations of solute the interaction effects are pretty small, so the dependence of vapor pressure on solute concentration remains roughly linear. The cool observation though is that at most temperatures and for most solvents, it doesn't matter what solute you use (as long as the solute itself doesn't have a vapor pressure), the vapor pressure of the solvent is still decreased by adding solute (which indicates that the entropic contribution is the most important part, and the interactions don't play a big role).
Now to sum up: for a given concentration of salt dissolved in water, there are more states accessible to the water molecules in the liquid phase than there are in pure water. So at every water temperature as we pour in energy to make it boil, there will be a lower vapor pressure than there would have been without the salt, and thus we won't get to the boiling point until the water has reached a higher temperature (until we've poured in more energy than we would have had to). Salt does disrupt the network of hydrogen bonds in the water molecules, but the effect isn't very big at reasonable concentrations of salt, and it's never big enough to counteract the entropic effect.
I have read that true steam is clear (transparent) water vapor. According to this theory, the white "steam" you see is really a small cloud of condensed water vapor droplets, a fine mist in effect. So what you are seeing is not more steam, but more condensation and more mist. The speed with which the steam/vapor/mist rises and disperses may also change.
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Best Answer
It depends on how much salt you add. Adding salt increases the boiling temperature but decreases the heat capacity, so it takes less heat for the solution to boil. For small amounts of salt, the total time increases; for larger amounts, it decreases.
Using simple chemistry, we can estimate how much longer boiling would take. The elevation in boiling temperature caused by adding the salt can be expressed as $$\Delta T=Kib$$ where $K\approx 0.5 \,\mathrm{K \,kg \,mol^{-1}}$ for water is called the ebullioscopic constant, $i\approx 2$ for salt ($\mathrm{NaCl}$ dissociates in water) and $b$ is the molality, that is, the number of moles of salt per mass of water.
Heat is absorbed according to the law $$Q = M C (T_f-T_i)$$ where $M$ is the mass, $C$ is the heat capacity, and $T_i$ and $T_f$ are the initial and final temperatures, respectively. The heat capacity of the solution of water + salt as a function of concentration (by mass) can be found here. Note that adding salt lowers the heat capacity.
Let $m_w$ be the mass of the water and $m_s$ be the mass of salt. $C_w \approx 4200\,\mathrm{J\, kg^{-1}\, K^{-1}}$ is the heat capacity of water, and $C_{sol}$ the heat capacity of the solution. To boil, the water without salt needs to absorb heat in the amount of $$Q_1 = m_w C_w (T_0-T_i)$$ The solution of water + salt has more mass and needs to reach a higher temperature, so it needs an amount of heat equal to $$Q_2 = (m_w+m_s) C_{sol} (T_0+\Delta T-T_i)$$
Assuming a constant heat source, the time required to heat to a certain temperature is proportional to the amount of heat needed. So if $t_1$ is the time it takes for pure water to boil and $t_2$ is the time it takes for water + salt to boil, the ratio of these two is $$\frac{t_2}{t_1}=\frac{Q_2}{Q_1}=\frac{C_{sol}}{C_w}\left(1 + \frac{m_s}{m_w}\right)\left(1+ \frac{\Delta T}{T_0-T_i}\right)$$
Now suppose we use $1\,\mathrm{kg}$ of water and start at $T_i=20^\circ \mathrm{C}$. Say we add $10 \,\mathrm{g}$ of salt; we can use this online calculator to obtain the number of moles, which in this case is $0.17\,\mathrm{mol}$. The changes in $C$ and $m$ are negligible for such a small amount of salt, so we get $$\frac{t_2}{t_1}=1.01$$ Thus, it takes about $1\%$ more time if you add a small amount of salt.
Say instead we add $50\,\mathrm{g}$ of salt, corresponding to about $0.86\,\mathrm{mol}$. Then according to the source above, the heat capacity of the solution is going to be $C_{sol}\approx 3900\,\mathrm{J\, kg^{-1}\, K^{-1}}$. Therefore, we get
$$\frac{t_2}{t_1}\approx \frac{3900}{4200} \times 1.05 \times \left( 1 + \frac{0.86}{80} \right) \approx 0.985$$
So if we add a larger amount of salt, the lowering in heat capacity is sufficient to offset the increase in boiling temperature. The solution of water + salt heats faster.
Note however that these changes are tiny. As pointed out in the comments, the non-linearity of the boiling temperature elevation law might affect the validity of results for larger concentrations. Moreover, hard-to-account-for effects caused by the change in density and viscosity might be relevant.