Introduction (Excel spreadsheet for calculation)

If you want to cool a warm drink relatively quickly in summer, you usually use ice cubes. The disadvantage, however, is that soft drinks are usually verwässern by the melting of the ice cubes. So wouldn’t it make sense to develop ice cubes that don’t melt? Or couldn’t one simply use copper cubes cooled in the freezer?

In order to answer this question, it will first be shown how the final temperature after complete melting of the ice cubes and mixing with the beverage can be calculated. For simplicity, it is assumed that heat is only transferred between the beverage and the ice cubes. In reality, the surroundings and the glass are also involved in the heat transfer that takes place, but this is neglected in the following.

The entire cooling process can basically be divided into three steps, which will be discussed in more detail below:

1. Heating the ice cubes to melting temperature
2. Melting of the ice cubes
3. Mixing the melted ice with the drink

For all those who do not want to deal with the derivation of the formulas, the final temperature can be calculated directly using the following Excel spreadsheet:

Heating the ice cubes to melting temperature

Ice cubes from the freezer usually have a temperature of about -20 °C. If the ice cubes are added to the beverage, they first heat up to the melting temperature of 0 °C. Using the specific heat capacity of the ice of c_ice = 2.0 kJ/(kg⋅K) and the mass m_ice = 50 g, the required heat Q_ice for this heating process can be determined based on the temperature rise of ΔT_ice = 20 °C:

\begin{align}
&Q_\text{ice} = c_\text{ice} \cdot m_\text{ice} \cdot \Delta T_\text{ice} \\[5px]
&Q_\text{ice} = 2.0 \tfrac{\text{kJ}}{\text{kg⋅K}} \cdot 0.05 \text{ kg} \cdot 20 \text{ °C} = 2 \text{ kJ}\\[5px]
\end{align}

To heat the ice cubes to melting temperature, 2 kJ of heat must be added. This required heat is supplied by the beverage. This means that 2 kJ of heat energy is drawn from the beverage to heat the ice cubes. The temperature of the beverage obviously drops as a result. The amount by which the temperature of the beverage is changed (ΔT₁) due to the release of heat Q_ice can be determined using the specific heat capacity of the beverage c_bev and the mass of the beverage m_bev. Since most beverages or soft drinks are almost entirely water, the specific heat capacity of water can be used (c_bev = 4.2 kJ/(kg⋅K)):

\begin{align}
&Q_\text{ice} = c_\text{bev} \cdot m_\text{bev} \cdot \Delta T_\text{1} \\[5px]
&\Delta T_\text{1} = \frac{Q_\text{ice}}{c_\text{bev} \cdot m_\text{bev}} \\[5px]
&\Delta T_\text{1} = \frac{2 \text{ kJ}}{4.2 \frac{\text{kJ}}{\text{kg⋅K}} \cdot 0.5 \text{ kg}} \approx 1 \text{ °C} \\[5px]
\end{align}

Due to the heating of the ice cubes to melting temperature, the temperature of the beverage decreases by 1 °C. As the temperature of the beverage was 20 °C at the beginning, it now has a temperature of 19 °C.

Melting of the ice cubes

When the ice cubes are heated to melting temperature, they begin to melt. During melting so-called latent heat of fusion Q_f is necessary to break the intermolecular bonds. This latent heat of fusion can be determined using the specific latent heat of fusion of ice with q_f = 334 kJ/kg and the mass of the ice cubes to be melted of m_ice = 50 g:

\begin{align}
&Q_\text{f} = q_\text{f} \cdot m_\text{ice} \\[5px]
&Q_\text{f} = 334 \frac{\text{kJ}}{\text{kg}} \cdot 0,05 \text{ kg} = 16.7 \text{ kJ} \\[5px]
\end{align}

To completely melt the ice cubes, 16.7 kJ of heat energy is required. This amount of heat is also supplied by the beverage. Thus, 16.7 kJ of heat energy is drawn from the beverage during melting. This is significantly more than during the heating of the ice cubes. Accordingly, the associated temperature change of the beverage (ΔT₂) is also higher:

\begin{align}
&Q_\text{f} = c_\text{bev} \cdot m_\text{bev} \cdot \Delta T_2 \\[5px]
&\Delta T_2 = \frac{Q_\text{f}}{c_\text{bev} \cdot m_\text{bev}} \\[5px]
&\Delta T_2 = \frac{16.7 \text{ kJ}}{4.2 \frac{\text{kJ}}{\text{kg⋅K}} \cdot 0.5 \text{ kg}} \approx 8 \text{ °C} \\[5px]
\end{align}

Due to the melting of the ice cubes, the temperature of the beverage decreases by 8 °C. Since the temperature of the beverage before melting was 19 °C, it now has a temperature of 11 °C.

Mixing the melted ice with the beverage (final temperature)

Although mixing of the 0 °C cold melted water with the beverage takes place at the same time as the ice cubes melt, this process can be considered separately from an energy point of view. This means that first the ice cubes are completely melted, as considered in the previous section. Afterwards the mixing between 0 °C cold melted water (T_w) and 11 °C cold beverage (T_bev) is considered. The final temperature of the mixture can be calculated with the Richmann’s law of mixtures:

\begin{align}
\boxed{T_\text{f}  = \frac{c_\text{bev} \cdot m_\text{bev} \cdot T_\text{bev} + c_\text{w} \cdot m_\text{w} \cdot T_\text{w}}{c_\text{bev} \cdot m_\text{bev} + c_\text{w} \cdot m_\text{w}}} ~~~\text{Richmann’s law} \\[5px]
\end{align}

For the specific heat capacity of the melted ice cubes c_w, as for the beverage, the specific heat capacity of water with c_w= 4.2 kJ/(kg⋅K) is used (in this case the specific heat capacities cancel each other out in the above formula). The mass of the melted water m_w is equal to the mass of the ice cubes with m_w = 50 g. The final mixing temperature T_f can thus be determined as follows:

\begin{align}
\require{cancel}
&T_\text{f}  = \frac{\cancel{c_\text{bev}} \cdot m_\text{bev} \cdot T_\text{bev} + \cancel{c_\text{w}} \cdot m_\text{w} \cdot T_\text{w}}{\cancel{c_\text{bev}} \cdot m_\text{bev} + \cancel{c_\text{w}} \cdot m_\text{w}} \\[5px]
&T_\text{f}  = \frac{ m_\text{bev} \cdot T_\text{bev} + m_\text{w} \cdot T_\text{w}}{m_\text{bev} + m_\text{w}} \\[5px]
&T_\text{f}  = \frac{ 0.5 \text{ kg} \cdot 11 \text{ °C} + 0.05 \text{ kg} \cdot 0 \text{ °C}}{0.5 \text{ kg} + 0.05 \text{ kg}} = 10 \text { °C} \\[5px]
\end{align}

The mixing thus causes the temperature of the beverage to drop again by ΔT₃ = 1 °C to a total of 10 °C.

Conclusion

If we look at the cooling of the beverage, which was divided into three steps, the following temperature change of the beverage can be seen in the respective cases:

• Temperature change due to the heating the ice cubes: ΔT₁ = 1 °C
• Temperature change due to the melting of the ice cubes: ΔT₂ = 8 °C
• Temperature change due to mixing of the melted water: ΔT₃ = 1 °C

Obviously, the largest temperature change is due to the melting of the ice cubes. This is because of the relatively large heat of fusion of water compared to the small amounts of heat transferred during the heating of the ice cubes or mixing. It is therefore essential for the cooling effect that the ice cubes melt! In this respect, a compromise must always be made between the desired cooling and the watering down of the beverage by the melted ice cubes.

If one were to use the aforementioned copper cubes instead of ice cubes, which one cools to -20 °C in the freezer, one would only obtain a final temperature of the beverage of 16.7 °C. Copper obviously does not melt when added to a beverage. It is only a “mixing” that can be described by Richmann’s law of mixtures. The copper cubes have a specific heat capacity of 0.385 kJ/(kg⋅K) and a mass of 0.488 kg (with identical volume to the 50-gram ice cubes):

\begin{align}
\require{cancel}
&T_\text{f}  = \frac{c_\text{bev} \cdot m_\text{bev} \cdot T_\text{bev} + c_\text{co} \cdot m_\text{co} \cdot T_\text{co}}{c_\text{bev} \cdot m_\text{bev} + c_\text{co} \cdot m_\text{co}} \\[5px]
&T_\text{f}  = \frac{ 0.5 \text{ kg} \cdot 4.2 \frac{\text{kJ}}{\text{kg⋅K}} \cdot 20 \text{ °C} + 0.488 \text{ kg} \cdot 0.385 \frac{\text{kJ}}{\text{kg⋅K}} \cdot (-20 \text{ °C})}{0.5 \text{ kg} \cdot 4.2 \frac{\text{kJ}}{\text{kg⋅K}} + 0.488 \text{ kg} \cdot 0.385 \frac{\text{kJ}}{\text{kg⋅K}}} \\[5px]
&T_\text{f}  = 16.7 \text{ °C} \\[5px]
\end{align}

Cooking on Mount Everest

With increasing altitude above sea level, the air pressure decreases more and more (see also the article on barometric formula). This shows the phenomenon that water begins to boil at significantly lower temperatures than one is used to at lower altitudes. At sea level at a pressure of 1.013 bar, water begins to boil at a temperature of 100 °C.

However, on Mount Everest at an altitude of 8849 m, the air pressure is only around 0.325 bar. Due to this significantly reduced pressure, the water already begins to boil at a temperature of around 71°C. However, since the temperature does not rise any further during boiling, the cooking of foods such as potatoes or pasta thus takes significantly longer (see also the article Why does the temperature remain constant during a change of state?).

Explanation with the particle model

The fact that the boiling point depends on the ambient pressure applies not only to water, but ultimately to all liquids. In particular, it is true that the boiling point decreases with decreasing pressure. This phenomenon can be explained qualitatively with the particle model of matter.

During boiling, the liquid vaporizes and becomes gaseous. In this vaporization process, energy is absorbed by the liquid and added to the molecules, allowing them to break free from the molecular binding forces of the liquid and enter the gas phase. At an ambient air pressure of 1 bar, vaporization of water takes place at a temperature of 100 °C.

However, if the ambient air pressure is increased, the air molecules collide more strongly with the surface of the liquid. In the process, the air molecules push the liquid molecules back into the liquid, so to speak. It thus becomes more difficult for the molecules in the liquid to pass into the gas phase. The water molecules consequently require greater energy and thus a higher temperature in order to escape the liquid phase. For this reason, with increased ambient air pressure, a higher boiling temperature is required to vaporize a liquid or bring it to a boil.

The boiling temperature of a liquid increases with increasing ambient pressure!

Increase of boiling temperature at elevated ambient pressure (pressure cooker)

Under high ambient pressure, water consequently also boils at higher temperatures. This is used, for example, in so-called pressure cookers to heat the water to over 100 °C. A pressure cooker seals the pot of water gas-tight. During vaporization, water normally expands 1700 times. However, since this is not possible with a sealed pot, the pressure consequently increases. A pressure relief valve usually limits the pressure to a maximum of 2 bar. The boiling temperature rises to around 120 °C at this increased pressure. As a result, food prepared in the pot is no longer cooked at just 100 °C, but at 120 °C!

Decrease of boiling temperature at reduced ambient pressure

If an increase in the ambient pressure leads to an increase in the boiling temperature, then in the opposite case this means that a decrease in the ambient pressure results in a decrease in the boiling temperature. And this is exactly what explains why water on Mount Everest boils at already 71 °C due to the lower pressure of only 0.325 bar. Preparing food that normally requires a temperature of 100 °C in water is therefore not so easy at high altitudes. At this point, one would have to use the pressure cooker already explained to obtain increased pressure and raise the boiling temperature.

The following experiment provides an impressive demonstration of the decrease in boiling temperature with decreasing pressure. For this purpose, a glass with water is placed under a vacuum chamber. A thermometer is placed in the glass to observe the temperature. The thermometer indicates a temperature of 20 °C. Now the vacuum pump is switched on and thus the pressure is reduced step by step. Below a pressure of about 0.023 bar, one then observes small bubbles rising in the water. This is the typical phenomenon when water boils, with the temperature still at 20 °C. And indeed, at a pressure of 0.023 bar, the water already begins to boil at 20 °C.

Increase in volume during vaporization

In the article Specific latent heat of vaporization, it was explained that the heat of vaporization is necessary for breaking the intermolecular bonds. In fact, however, when a liquid vaporizes, it is not only the bindung energies that must be overcome. In general, a substance also expands greatly during vaporization. Steam, for example, occupies a volume almost 1700 times greater than liquid water at the same pressure of 1 bar. Thus, one liter of liquid water becomes about 1700 liters of water vapor after vaporization!

Energy is also required for this expansion of the volume of a substance against the acting ambient pressure (displacing the vapor into the environment). This energy to be added is also supplied in terms of heat and is already taken into account in the heat of vaporization! In most cases, vaporization takes place in an open system. The vaporizing liquid can thus expand freely at constant ambient pressure. This is also referred to as isobaric vaporization. In this case, the energy W required for the change in volume can be determined from the product of pressure p and volume change ΔV:

\begin{align}
&\boxed{W = p \cdot \Delta V} \\[5px]
\end{align}

This energy is also referred to as pressure-volume work or displacement work due. Note, however, that this energy to be added is not supplied in the form of mechanical work, but also as heat. So don’t be misled by the term work.

The amount of energy required to generate the gas volume during vaporization can be thought of as the expansion of a balloon. Energy is required to inflate the balloon. In a figurative sense, this corresponds to the generation of the gas volume against the acting ambient pressure.

Enthalpy of vaporization

Consequently, the heat of vaporization can be divided into a portion that causes the change in internal energy ΔU (in terms of changed binding energies), and a portion that results in the change of volume:

\begin{align}
\label{h}
&\boxed{Q_\text{V} = \Delta U + p \cdot \Delta V }:= \Delta H_\text{V} ~~~\text{enthalpy of vaporization} \\[5px]
\end{align}

The sum of internal energy and the product of pressure and volume is also called enthalpy. Equation (\ref{h}) thus gives the change in the enthalpy of the substance due to an isobaric supply of heat during vaporization. The heat of vaporization is therefore also called enthalpy of vaporization.

In the case of isobaric vaporization, the added heat of vaporization completely benefits the change of enthalpy and is therefore also called enthalpy of vaporization!

Note that the heat of vaporization and the enthalpy of vaporization are identical in value, but have different thermodynamic meanings! The enthalpy is a state variable which describes the energetic state of a substance on the basis of the internal energy and the pressure and volume (all three variables are state variables!). Heat, on the other hand, is a process quantity that only describes the process of energy transfer as heat into or out of a thermodynamic system. The heat of vaporization transferred across the system boundary is the cause of the change in enthalpy of the system.

Enthalpy of vaporization using the example of water

For water, the pressure-volume work during vaporization is to be determined in the following. At first, 1 kilogram of liquid water at a pressure of 1 bar occupies a volume of about 1 liter. After complete vaporization, the volume has increased to about 1692 liters (this value can be determined using the Van der Waals equation). The change in volume is therefore ΔV = 1691 liters. Consequently, an energy of around 169 kJ is required to change the volume:

\begin{align}
&W = p \cdot \Delta V = 1 \cdot 10^5 \tfrac{\text{N}}{\text{m²}} \cdot 1.691 \text{ m³} \approx 169 \text{ kJ} \\[5px]
\end{align}

According to the literature, the total heat of vaporization to be added for 1 kilogram of water is 2257 kJ. Thus, at an ambient pressure of 1 bar, around 7.5 % of the total heat of vaporization is used increase the volume during vaporization. The remaining 92.5% of the heat of evaporation is then actually used to change the binding energy (change in internal energy).

If growing fruits are exposed to sub-zero temperatures (frost) on cold spring nights, they are in danger of freezing. The harvest of entire vineyards or other plantations is in danger of failing. As paradoxical as it may seem at first, the solution is the freezing of the plants by sprinkling with water!

Sprinklers for frost protection do not use warm water for any heating of the plants, but ordinary tap water. After all, sooner or later the warm water would freeze anyway. And in fact one uses just this freezing of the water for the frost protection of the sensitive plants!

The sprayed water mist forms a water film around the plant, which then cools down. The solidification process of the water finally starts at a temperature of 0°C. During the freezing of the water, latent heat of solidification is released (also known as heat of crystallization). Thus, although heat is drawn from the plant by the cold environment on the one hand, heat is simultaneously added to the plant by the released heat of crystallization on the other hand. This ultimately means that the temperature does not drop further during the freezing of water and remains constant at 0°C despite the surrounding subzero temperatures. The plant is thus protected from temperatures below zero, which would otherwise cause the cell water in the plant to freeze.

The following example illustrates the enormous amount of heat released when water freezes. During the freezing of 1 kilogram of water, a latent heat of solidification of 334 kJ is released. With this amount of heat, starting from room temperature, you could bring 1 kilogram of water to a boil! This example shows how much heat is released when water freezes and thus protects the plant by preventing the temperature from falling below 0°C.

Note that the freezing point of the cell water of plants (and also of most other living organisms) is -2 °C, which is somewhat below the freezing point of water anyway, due to the substances dissolved in it. The plants thus have a kind of natural frost protection. Since the temperature of the water freezing on the plant does not drop below 0 °C during, the cell water inside the plant remains liquid.

During sprinkling with water, the crops or plants remain protected from sub-zero temperatures due to the heat of solidification released during freezing!

Of course, this method only works as long as the solidification process continues and thus the release of heat of solidification is ensured, so that the temperature in the plant does not drop below 0 °C. Therefore, this method must be maintained by permanent sprinkling with water as long as there is a risk of the plant freezing due to frost. If, on the other hand, the sprinkling of water were interrupted, the water film would freeze completely. After complete freezing, the temperature of the plant would drop below 0 °C and the plant would eventually freeze to death!

To vaporize a liquid, energy as heat must be transferred to the substance in order to break the intermolecular bonds so that the substance becomes gaseous. In the case of pure substances, the temperature remains constant until the liquid has completely vaporized. The heat added during vaporization therefore does not result in an increase in temperature, since it is used to break the intermolecular bonds (hydrogen bond). The heat to be added for the complete vaporization of a certain amount of liquid is also referred to as heat of vaporization or, more generally, as latent heat.

Water requires a very large amount of heat to vaporize. For example, to vaporize 1 kg of water, a heat energy of 2257 kJ is required. If we compare this amount of heat with the heating of water from 20 °C to 100 °C, only 336 kJ is required. Thus, more than 6 times as much heat is needed for vaporization as was necessary for heating! The transferred heat of vaporization cannot simply have disappeared due to the conservation of energy. Rather, this enormous amount of energy is stored as internal energy in the gas phase.

Water needs a multiple of the amount of heat for vaporization compared to heating up to boiling temperature!

During condensation, i.e. when gaseous water liquefies on a cold object, the previously absorbed latent heat is released again. The emitted heat from the substance is absorbed by the cooler object. The (internal) energy of the water decreases and the intermolecular bonds can form again, resulting in the liquid state. In the case of condensation, one also speaks of heat of condensation, which is also a form of latent heat. The amount of heat of condensation is the same as the heat of vaporization.

The energy absorbed by a substance during vaporization in the form of heat of vaporization (latent heat) is released during condensation in the form of heat of condensation (latent heat)!

This understanding now also explains why steam burns are generally much more painful and dangerous than water burns. If the relatively cool skin comes into contact with water vapor (steam), the water condenses there and heat of condensation is released and transferred to the skin. As already explained, due to the large amount of latent heat involved, there is a huge amount of thermal energy transferred. Thus, in contact with steam, significantly more heat is transferred to our skin than in contact with liquid water, although the temperature is the same in both cases (100 °C).

Steam burns are more dangerous compared to water burns because additional latent heat is transferred in the form of heat of condensation!

Note that the human perception of warm or cold is not based on temperatures, but on heat flows (transferred heat per unit time). Thus, although the temperatures are identical at 100 °C in both cases, condensation results in a much larger heat flow. This greater heat flow not only causes a psychologically warmer perception, but also leads physically to more dangerous burns. More information on the perception of warm and cold can be found in the article Why does metal feel colder than wood.

The simple answer to this question would be: Water cools the fire site and thus extinguishes the fire. However, one could now argue: If the water poured on the fire becomes hot and vaporizes, then the cooling effect is no longer there. Thus, the vaporization of the water should be a disadvantage if you want to extinguish a fire – right? In fact, the opposite is true: In particular, the vaporization of water plays a central role in effectively extinguishing a fire! To understand this, let’s take a closer look at the processes involved in the vaporization of water.

To vaporize water, the water must first be heated to boiling temperature. Heating one kilogram of water from 20 °C to 100 °C requires 336 kJ of heat. For the subsequent vaporization, further heat energy must be added to break the intermolecular bonds that hold the water molecules together in the liquid phase. The complete breaking of these hydrogen bonds requires another 2257 kJ of heat. This heat energy required for vaporization is also known as the heat of vaporization. Thus, vaporizing water requires more than 6 times as much heat as was needed to heat it to boiling temperature.

Water needs a multiple of the amount of heat for vaporization compared to heating up to boiling temperature!

The very large heat of vaporization of water is the reason why water is particularly suitable for extinguishing a fire. When water is poured onto the fire site, the water is not only heated, but also vaporized. As explained, the water requires a great deal of heat specifically for vaporization. This energy is supplied by the fire site. This means that a great amount of heat is absorbed from the burning material by the vaporization of the water. As a result, the fire site cools down and the temperature drops. If the temperature finally falls below the ignition temperature of the burning substance, the fire goes out.

By absorbing a very large amount of heat during vaporization, water draws energy from the fire site and thus cools it down until the fire goes out!

If the water did not vaporize when pouring it on a fire, but only heated up, then the water would not draw the enormous heat energy from the fire site due to the lack vaporization. So the water would not have the cooling effect by far and could not extinguish the fire so effectively. The vaporization of the water is therefore essential when it comes to extinguishing a fire with water. The temperature of the water plays a negligible role. Boiling water is also suitable for firefighting (at least when it comes to fires that can be extinguished with water – water must not be used for metal fires, as it decomposes and forms highly explosive oxyhydrogen gas!).

In fact, the excellent extinguishing effect of water is not only due to the great amount of heat of vaporization. The fact that water expands strongly during vaporization also plays an important role. One liter of liquid water, at constant pressure, becomes about 1700 liters of water vapor. This enormous increase in volume during vaporization displaces the oxygen necessary for combustion from the fire site and the fire suffocates. This happens particularly effectively when the liquid water is atomized into droplets (water mist) and sprayed onto the fire site. In this case, even oil or grease fires can be extinguished with the water mist, which may not normally be extinguished with liquid water. This is because in this case the fine water droplets vaporize before they come into contact with the burning oil. This prevents the oil from being swept up into the air by vaporizing water, which normally leads to an increase in the fire rate.

Melting and solidification

In the article on specific latent heat of fusion, it was explained in detail that when a solid is to be melted, energy is required to break the bonds so that the substance changes to the liquid phase. This energy is added to the substance as heat and is called heat of fusion. In the case of pure substances, the temperature remains constant during melting.

If the liquid substance is to be solidified, then the heat previously added must be released so that the molecules can bind to each other again. In this case, the heat energy to be given off during the solidification of the liquid is called heat of solidification. Even while the heat of solidification is being released, the temperature remains constant. It only drops when the liquid substance has completely solidified. The temperature at which a substance solidifies corresponds to the melting point of the substance (also called solidification point).

For a more detailed explanation of the molecular processes that occur during solidification and the reason why the temperature remains constant, see article Why does the temperature remain constant during a change of state (phase transition)?

Specific heat of fusion and solidification

Due to the conservation of energy, the amount of heat absorbed by a substance to melt (heat of fusion) is equal to the amount of heat released by the substance during solidification (heat of solidification). For this reason, the specific heat of solidification q_s, as the ratio of the heat of solidification Q_s and the solidifying mass m_s, is as great as the specific heat of fusion q_f.

\begin{align}
&\boxed{q_\text{s} = \frac{Q_\text{s}}{m_\text{s}}}~~~\text{where}~~~\boxed{q_\text{s} = q_\text{f}} \\[5px]
\end{align}

In the article Specific latent heat of fusion, the specific heats of fusion or solidification are listed for selected substances. Using these values, the heat of solidification Q_s to be released for a given mass m_s to be solidified can be determined with the following formula:

\begin{align}
&\boxed{Q_\text{s} = m_\text{s} \cdot q_\text{s}}\\[5px]
\end{align}

In the case of solidification, again one speaks of latent heat, since the heat to be released during solidification is not directly evidenced by a change in temperature (from the Latin word “latere”, which means “to be hidden” or “not to appear directly”).

In the case of water, the latent heat to be released during solidification is 334 kJ per kilogram, which is about the same as the amount of heat that would be needed to bring the water to a boil, starting at room temperature. This is therefore a very large amount of heat that has to be released or that is given off by the liquid water during solidification. This explains, for example, why sprinkling plants with water at temperatures below freezing point is used as frost protection.

Process of melting

If a solid is heated more and more, then at some point the melting point is reached. At this point, the state of matter changes and the solid finally begins to melt. During melting, no further increase in temperature is observed for pure substances, despite the continued supply of heat energy. During melting, the energy obviously no longer benefits the increase of the vibrational energy of the molecules, which would otherwise mean an increase in temperature (see also article Temperature and particle motion).

During melting, the transferred energy leads to an increase in the internal energy in terms of changed binding energies between the molecules in the solid and liquid state. The intermolecular bonds in the solid state are being broken by the added heat energy, thus allowing the transition to the liquid state. In the liquid state, the molecules are less strongly bound to each other due to the lower binding forces. In this way, the molecules are no longer bound to a specific location. That is why liquids do not have a solid form.

During melting, energy must be added to break the intermolecular bonds. In the case of pure substances, the temperature of the solid remains constant until the process of melting is completed!

More detailed information on this can also be found in the article Why does the temperature remain constant during a change of state (phase transition)?

The question arises as to how much heat must be added in order to completely melt a certain amount of a solid. The heat required for this is also referred to as the heat of fusion or enthalpy of fusion. This heat of fusion does not include the amount of heat required to heat the solid to the melting point. The heat of fusion therefore only includes the heat energy to be added during melting if the solid has already been heated to melting temperature.

The heat of fusion (enthalpy of fusion) is the heat energy to be added to a solid at its melting point in order to completely melt a certain amount of the substance!

Since the heat of fusion added during melting is not directly noticeable in an increase in temperature, but can nevertheless be found in the molten substance in the form of internal energy, the heat of fusion is also referred to as latent heat. The term “latent” comes from Latin and means “to be hidden” or “not to appear directly”.

Experimental determination of the heat of fusion

Experimental setup

Using frozen water (ice) as an example, the heat of fusion required to melt a certain amount of ice will be determined experimentally in the following. For this purpose, ice from the refrigerator is placed in an electrically heated vessel. The vessel should be thermally insulated so that the melting of the ice is due solely to the power output of the heater (which can be easily determined) and not to the heat from the warmer surroundings, which would normally cause the ice to melt. Through a small hole at the bottom of the vessel, the melted water is collected on a balance. In this way the melting process over time is observed.

The added heat of fusion can be determined via the electrical power of the heater, which is completely converted into heat. The heat energy Q_f (= latent heat of fusion) added at a power P is obtained by the operating time t of the heater according to the following formula:

\begin{align}
\label{q}
Q_\text{f} = P \cdot t \\[5px]
\end{align}

Observation

First, the ice is heated to melting temperature with the heater. When the ice begins to melt, the experiment can be started at any time. To do this, the balance is reset to zero and the time measurement is started. The ice gradually melts and the melted mass is displayed on the balance. At regular time intervals, the displayed mass of the balance is recorded. At each time t, the added heat of fusion Q_f up to that point can be determined using formula (\ref{q}). In this way one obtains a statement which amount of heat led to the melting of which mass m_f.

Evaluation

If one plots the melted mass as a function of the added heat, then a proportional relationship becomes apparent. This means that in order to melt twice the amount of ice, twice the amount of heat must be added. The evaluation of the experimental data shows that about 35 kJ of heat are necessary to melt 100 g of water. With a supplied heat energy of about 70 KJ, twice the ice mass of 200 g has then finally melted.

Especially with regard to the comparability of the heats of fusion of different solids, it therefore makes sense to always relate the heats of fusion Q_f to a standardized amount of mass to be melted (e.g. 1 kilogram or 1 gram). This constant ratio between the heat of fusion and the mass m_f to be melted is called specific heat of fusion or specific enthalpy of fusion q_f:

\begin{align}
&\boxed{q_\text{f} = \frac{Q_\text{f}}{m_\text{f}}}~~~[q_\text{f}]=\frac{\text{J}}{\text{kg}}~~~~~\text{specific heat of fusion} \\[5px]
\end{align}

From the experiment, a specific heat of fusion of around q_f = 350 kJ/kg is finally obtained for ice. This means that 350 kJ of heat is required to melt 1 kilogram of ice. However, with the experimentally determined heat of fusion using the described experimental setup, it must be noted that the heat emitted by the heater does not completely benefit the melting of the ice. Some of the heat is also used to heat the vessel and is thus transferred to the surroundings. Therefore, a lower amount of heat is used for the melting of the ice than calculated with formula (\ref{q}). The literature value for the specific heat of fusion of ice is therefore somewhat lower with q_f = 334 kJ/kg.

Specific heat of fusion is the heat of fusion to be added per unit mass of a solid to be melted!

In the case of ice, the latent heat to be added for melting is 334 kJ per kilogram, which is about the same as the amount of heat that would be needed to bring the water to a boil from room temperature. A great deal of heat must therefore be used for melting. This explains, for example, why ice cubes are ideal for cooling drinks.

Conclusion

The specific heat of fusion q_f describes the relationship between the mass to be melted m_f and the heat of fusion to be added for this purpose Q_f:

\begin{align}
&\boxed{Q_\text{f} = q_\text{f} \cdot m_\text{f}} ~~~\text{heat of fusion} \\[5px]
\end{align}

Specific heat of fusion of selected substances

If the experiment described above were to be carried out with other solids instead of ice (e.g. metals that are being molten), it would be shown that these substances melt at different rates. As a result, more or less heat energy is required to melt a given mass of the respective substance. The specific latent heat of fusion is therefore dependent on the substance.

The greater the specific heat of fusion of a substance, the more heat is required to melt a certain mass. Solids with large specific heats of fusion therefore do not melt as quickly. The table below shows the specific heats of fusion of selected substances.

Note: The relatively high specific heat of fusion of aluminum is, among other things, a reason why the production of aluminum, especially the melting down, is very energy-intensive and thus expensive!

Vaporization and condensation

In the article on specific latent heat of vaporization, it was explained in detail that when a liquid is to be vaporized, energy is required to break the bonds so that the substance changes to the gaseous phase. This energy is added to the substance as heat and is called heat of vaporization. In the case of pure substances, the temperature remains constant during vaporization.

If the gaseous substance is to be liquefied, then the heat previously added must be released so that the molecules can bind to each other again. In this case, the heat energy to be given off during the condensation of the gas is called heat of condensation. Even while the heat of condensation is being released, the temperature remains constant. It only drops again when the gaseous substance has completely condensed. The temperature at which a substance condenses corresponds to the boiling point of the substance.

For a more detailed explanation of the molecular processes that occur during condensation and the reason why the temperature remains constant, see article Why does the temperature remain constant during a change of state (phase transition)?

Specific heat of vaporization and condensation

Due to the conservation of energy, the amount of heat absorbed by a substance to vaporize (heat of vaporization) is equal to the amount of heat released by the substance during condensation (heat of condensation). For this reason, the specific heat of condensation q_c, as the ratio of the heat of condensation Q_c and the condensing mass m_c, is as great as the specific heat of vaporization q_v.

\begin{align}
&\boxed{q_\text{c} = \frac{Q_\text{c}}{m_\text{c}}}~~~\text{where}~~~\boxed{q_\text{c} = q_\text{v}} \\[5px]
\end{align}

The specific heat of condensation depends on the substance and indirectly on the pressure, since the boiling point (condensation point) is influenced by the pressure. In the article Specific latent heat of vaporization, the specific heats of vaporization or condensation are listed for selected substances. Using these values, the heat of condensation Q_c to be released for a given mass m_c to be condensed can be determined with the following formula:

\begin{align}
&\boxed{Q_\text{c} = m_\text{c} \cdot q_\text{c}}\\[5px]
\end{align}

In the case of condensation, again one speaks of latent heat, since the heat to be released during condensation is not directly evidenced by a change in temperature (from the Latin word “latere”, which means “to be hidden” or “not to appear directly”). In the case of water, the latent heat to be released during condensation is 2257 kJ per kilogram. This is more than five times the amount of heat that would have been required to heat the water from 0 °C to 100 °C! This explains, for example, why steam burns are more dangerous than water burns.

The process of releasing the heat of condensation

Whereas in the case of vaporization one usually has a very clear idea of the process of the transfer of heat, one sometimes has a little difficulty with the releasing of heat in the case of condensation. If, for example, water is vaporized in a pot, then the process of transferring heat takes place actively, so to speak, by heating on a hotplate. If, on the other hand, the gaseous water condenses on a glass plate, for example, then this process does not appear to have been actively forced. It seems as if the process would run by itself, without the need to actively dissipate heat.

In fact, however, heat must also be released actively in the process. In this case, this is done by the cooler glass plate, which absorbs the heat of condensation and thus dissipates it to the surroundings. In the same way as the hotplate plate actively transfers heat (of vaporization) to the water, the cold glass plate also actively absorbs heat (of condensation).

If, for example, the glass plate had the same temperature of 100 °C as the water vapor, then no more water would condense in this case either, since there would be no heat flow from the vapor to the glass, i.e. no absorption of heat. Conversely, if the hotplate had the same temperature of 100 °C as the liquid water, then vaporization would not occur in this case either, since there is no temperature difference to drive the heat flow.

Process of vaporization

If a liquid is heated more and more, then at some point the boiling point is reached. At this point, the state of matter changes and the liquid finally begins to vaporize (also referred to as boiling). During vaporization, no further increase in temperature is observed for pure substances, despite the continued supply of heat energy. During vaporization, the energy obviously no longer benefits the increase of the kinetic energy of the molecules, which would otherwise mean an increase in temperature (see also article Temperature and particle motion).

During vaporization, the transferred energy leads to an increase in the internal energy in terms of changed binding energies between the molecules in the liquid and gaseous state. The intermolecular bonds in the liquid state are being broken by the added heat energy, thus allowing the transition to the gaseous state. In the gaseous state, the molecules are only relatively weakly bonded to each other due to the lower binding forces.

During vaporization, energy must be added to break the intermolecular bonds. In the case of pure substances, the temperature of the liquid remains constant until the process of vaporization is completed!

More detailed information on this can also be found in the article Why does the temperature remain constant during a change of state (phase transition)?

The fact that heat has to be supplied permanently to drive the breaking of intermolecular bonds is evident, for example, in the boiling of water. If water is boiled in a pot, the water only vaporizes as long as the hotplate remains switched on. However, if the heat supply is interrupted, the water also stops boiling.

The question arises as to how much heat must be added in order to completely vaporize a certain amount of a liquid. The heat required for this is also referred to as the heat of vaporization or enthalpy of vaporization. This heat of vaporization does not include the amount of heat required to heat the liquid to the boiling point. The heat of vaporization therefore only includes the heat energy to be added during vaporization if the liquid has already been heated to boiling temperature.

The heat of vaporization (enthalpy of vaporization) is the heat energy to be added to a liquid at its boiling point in order to completely vaporize a certain amount of the substance!

For the difference between the terms heat and enthalpy, see the article Difference between latent heat of vaporization and enthalpy of vaporization.

Since the heat of vaporization added during vaporization is not directly noticeable in an increase in temperature, but can nevertheless be found in the vaporized substance in the form of internal energy, the heat of vaporization is also referred to as latent heat. The term “latent” comes from Latin and means “to be hidden” or “not to appear directly”.

Experimental determination of the heat of vaporization

Experimental setup

Using water as an example, the heat of vaporization required to vaporize a certain amount of water is to be determined experimentally in the following. For this purpose, water is first heated to boiling temperature with an immersion heater. Then, the vaporization of the water mass over time is observed using a balance on which the experimental setup is placed.

The added heat of vaporization can be determined via the electrical power of the immersion heater, which is completely converted into heat. The heat energy Q_v (= heat of vaporization) added at a power P is obtained by the operating time t of the immersion heater according to the following formula:

\begin{align}
\label{q}
Q_\text{v} = P \cdot t \\[5px]
\end{align}

Observation

First, the water is heated to boiling temperature with the immersion heater. When the water begins to vaporize, the experiment can be started at any time. To do this, the balance is reset to zero and the time measurement is started. The water gradually vaporizes and the vaporized mass is displayed on the balance. At regular time intervals, the displayed mass of the balance is recorded. At each time t, the added heat of vaporization Q_v up to that point can be determined using formula (\ref{q}). In this way one obtains a statement which amount of heat led to the vaporization of which mass m_v.

Evaluation

If one plots the vaporized mass as a function of the added heat, then a proportional relationship becomes apparent. This means that in order to vaporize twice the amount of water, twice the amount of heat must be added. The evaluation of the experimental data shows that about 120 kJ of heat are necessary to vaporize 48 g of water. With a supplied heat energy of about 240 KJ, twice the water mass of 96 g has then finally vaporized.

Especially with regard to the comparability of the heats of vaporization of different liquids, it therefore makes sense to always relate the heats of vaporization Q_v to a standardized amount of mass to be vaporized (e.g. 1 kilogram or 1 gram). This constant ratio between the heat of vaporization and the mass m_v to be vaporized is called specific heat of vaporization or specific enthalpy of vaporization q_v:

\begin{align}
&\boxed{q_\text{v} = \frac{Q_\text{v}}{m_\text{v}}}~~~[q_\text{v}]=\frac{\text{J}}{\text{kg}}~~~~~\text{specific heat of vaporization} \\[5px]
\end{align}

From the experiment, a specific heat of vaporization of around 2500 kJ/kg is finally obtained for water. This means that 2500 kJ of heat is required to vaporize 1 kilogram of water. However, with the experimentally determined heat of vaporization using the described experimental setup, it must be noted that the heat emitted by the immersion heater does not completely benefit the vaporization of the water. Some of the heat is also used to heat the vessel and is thus transferred to the surroundings. Therefore, a lower amount of heat is used for the vaporization of the water than calculated with formula (\ref{q}). The literature value for the specific heat of vaporization of water is therefore somewhat lower with 2257 kJ/kg.

Specific heat of vaporization is the heat of vaporization to be added per unit mass of a liquid to be vaporized!

Conclusion

The specific heat of vaporization q_v describes the relationship between the mass to be vaporized m_v and the heat of vaporization to be added for this purpose Q_v:

\begin{align}
&\boxed{Q_\text{v} = q_\text{v} \cdot m_\text{v}} ~~~\text{heat of vaporization} \\[5px]
\end{align}

In the case of water, the heat of vaporization to be added is more than five times as great as the amount of heat that would have had to be used to heat the water from 0 °C to 100 °C. This relatively large heat of vaporization is one of the reasons why a fire is extinguished excellently with water.

Specific heat of vaporization of selected substances

If the experiment described above is carried out with a water-alcohol mixture or with other liquids instead of pure water (e.g. molten metals that are vaporized), it can be seen that these substances vaporize at different rates. Consequently, more or less heat energy is required to vaporize a given mass of the substance. The specific heat of vaporization is therefore dependent on the substance.

The greater the specific heat of vaporization of a substance, the more heat is required to vaporize a given mass. Substances with large specific heats of vaporization therefore do not vaporize as quickly. The table below shows the specific heats of vaporization of selected liquids.

It should be noted that the specific heats of vaporization are indirectly influenced by the ambient pressure, since this changes the boiling temperatures (see also the article Why does water boil faster at high altitudes?)! However, since most liquids are vaporized at an ambient pressure of 1 bar, the specific heat of vaporization usually refers to the boiling temperature at 1 bar.

Note that the temperature of not all liquids remains constant during vaporization! For example, petroleum as a mixture of different substances does not vaporization at a boiling point, but within a boiling range. In the case of petroleum, this boiling range is between 180 °C and 330 °C. The heat added during the phase transition is therefore used both to raise the temperature and to drive the vaporization. It is therefore not possible to determine exactly what proportion of the heat added is used for the temperature increase or for the vaporization. Consequently, no (specific) heat of vaporization can be assigned to such a mixture of substances. In general, such a boiling range occurs with mixtures of substances, whereas pure substances usually have a boiling point.