91 Latent Heats

Latent Heats

We have learned that evaporation of liquid molecules transfers thermal energy from the liquid to the air. The heat of vaporization quantifies how much energy is transferred. The reverse process of , in which vapor molecules re-enter a liquid or stick together on a surface, will bring thermal energy out of the air. The  also quantifies the energy transferred during condensation. Changing from solid to liquid, known as , requires energy just as evaporation does. In similar fashion to vaporization, the amount of energy removed by melting is quantified by a latent heat, in this case the (Lf). For water, Lf = 334 kJ/kg, which means that for every kilogram of ice melted, 334 kiloJoules of energy input is needed. Freezing releases the same amount of energy into the environment that melting requires as input.


Figure a shows a four by four square lattice object labeled solid. The lattice is made of four rows of red spheres, with each row containing four spheres. The spheres are attached together horizontally and vertically by springs, defining vacant square spaces between the springs. A short arrow points radially outward from each sphere. The arrows on the different spheres point in different directions but are the same length, and one of them terminates at a dashed circle that is labeled limits of motion. To the right of this object are shown two curved arrows. The upper curved arrow points rightward and is labeled “energy input” and “melt.” The lower arrow points leftward and is labeled “energy output” and “freeze.” To the right of the curved arrows is a drawing labeled liquid. This drawing contains nine red spheres arranged randomly, with a curved arrow emanating from each sphere. The arrows are of different lengths and point in different directions.Figure b shows a drawing labeled liquid that is essentially the same as that of figure a. To the right of this drawing are shown two curved arrows. The upper curved arrow points rightward and is labeled “energy input” and “boil.” The lower arrow points leftward and is labeled “energy output” and “condense.” To the right of the curved arrows is another drawing of randomly arranged red spheres that is labeled gas. This drawing contains eight red spheres and each sphere has a straight or a curved arrow emanating from it. Compared to the drawing to the left that is labeled liquid, these arrows are longer and the red spheres are more widely spaced.
Energy is required to partially overcome the attractive forces between molecules in a solid to form a liquid. That same energy must be removed for freezing to take place. (b) Molecules are separated by large distances when going from liquid to vapor, requiring significant energy to overcome molecular attraction. The same energy must be removed for condensation to take place. When causing a phase change by adding or removing thermal energy, there is no temperature change until the phase change is complete. Image Credit: OpenStax CNX


Just as with evaporation, melting is a spontaneous process. We don’t actively control the behavior of the molecules, but they simply trend toward a configuration which maximizes the entropy. The temperatures at which phase changes will lead to an increase in entropy depends on the pressure of the surrounding air. The temperature at which melting occurs (under a given pressure) is known as the melting point. The melting and boiling points and latent heats of various substances at standard atmospheric pressure are shown in the following chart:

Melting and Boiling points, and Latent Heats at Standard Atmospheric Pressure[1][/footnote]
Lf Lv
Substance Melting point (ºC) kJ/kg kcal/kg Boiling point (°C) kJ/kg kcal/kg
Helium −269.7 5.23 1.25 −268.9 20.9 4.99
Hydrogen −259.3 58.6 14.0 −252.9 452 108
Nitrogen −210.0 25.5 6.09 −195.8 201 48.0
Oxygen −218.8 13.8 3.30 −183.0 213 50.9
Ethanol −114 104 24.9 78.3 854 204
Ammonia −75 108 −33.4 1370 327
Mercury −38.9 11.8 2.82 357 272 65.0
Water 0.00 334 79.8 100.0 22565 5396
Sulfur 119 38.1 9.10 444.6 326 77.9
Lead 327 24.5 5.85 1750 871 208
Antimony 631 165 39.4 1440 561 134
Aluminum 660 380 90 2450 11400 2720
Silver 961 88.3 21.1 2193 2336 558
Gold 1063 64.5 15.4 2660 1578 377
Copper 1083 134 32.0 2595 5069 1211
Uranium 1133 84 20 3900 1900 454
Tungsten 3410 184 44 5900 4810 1150

Reinforcement Exercises

Everyday Example: Entropy Change of Melting

If you place ice in a  warm room and leave it alone, it will melt. The ice would be a because it will happen all on it’s own, so we should find that melting increases the total entropy (ΔS > 0). Let’s check that out. We’ll keep it simple and calculate change in entropy of one kilogram of ice, which melts at 0 °C, or 273 K.

    \begin{equation*} \Delta S_{ice} = Q/T = \frac{mL_f}{T} = \frac{(1\,\bold{kg})(334,000\,\bold{J/kg})}{273 \,\bold{K}} = 1223\, \bold{J/K} \end{equation*}

Next we calculate the change in for the room. The same that went into melting the ice came out of the room, so the Q for the room is the same as for the ice, only negative. Let’s pick a typical room temperature of 20 °C for our example, which is 293 K:

    \begin{equation*} \Delta S_{room} = Q/T = \frac{-mL_f}{T} = \frac{-(1\,\bold{kg})(334,000\,\bold{J/kg})}{293 \,\bold{K}} = -1140\, \bold{J/K} \end{equation*}

Now we just add up the two changes to get the total:

    \begin{equation*} \Delta S_{T} =\Delta S_{ice} + \Delta S_{room} = 1223\, \bold{J/K} + (-1140)\, \bold{J/K} = 17\, \bold{J/K} \end{equation*}

Our entropy change is greater than zero, so the Second Law states that ice melting in warm room is a , which we observe it to be.

Everyday Examples: Melting Ice for Drinking Water

Spending more than a day in frozen landscapes (high altitude,  high latitude, or both), will require ice/snow to make drinking water.  Let’s determine the thermal energy required to melt 10 kg of  ice that started at 0 °C.

First we look up the for water in the previous chart and find Lf  = 334 kJ/kg, or 334,000 J/kg. Using this value in the latent heat formula:

    \begin{equation*} Q = mL_f  = (10\,\bold{kg})(334,000\,\frac{\bold{J}}{\bold{kg}})=3,340,000 \,\bold{J} \end{equation*}

Now that we are familiar with the idea of latent heat, let’s combine that with our understanding of the energy required to change temperature, to get a full of the minimum amount of stove fuel required for a three-person, round-trip expedition to the summit of Denali, North America’s highest mountain, at 20,380 ft (6495 m). Allowing for gradual acclimatization to altitude will reduce the likelihood of acquiring high altitude pulmonary edema and/or high altitude cerebral edema, so we will plan 21 days on the mountain.

First, let’s figure out how much water we need. The combination of acclimatization and climbing effort will require at least 4 liters of water per person per day. The total we need is then:

    \begin{equation*} V = (21 days)(3 people)(4\,\frac{\bold{L}}{day \cdot person}) = 252 \,\bold{L} \end{equation*}

The of water is 1 kg/L so we will also need 252 kg of water.  To get that of water we will need to warm and melt that mass of snow each day.  The average temperature on the mountain will be -20 °C and we will assume the snow to be melted starts off at this temperature. Let’s find the energy required to warm that mass of -20 °C snow up to 0 °C. We use a chart of specific heats to find cice = 2 kJ/kg = 2,000 J/kg. Using that value in our equation relating heat and temperature change:

    \begin{equation*} Q = mc\Delta T = (252\, \bold{kg})(2,000\,\bold{\frac{J}{kg\, C^{\circ}}})(20\,\frac{\bold{C^{\circ}}}) = 10,080,000 \,\bold{J} \end{equation*}

Next we need the energy to melt the snow:

    \begin{equation*} Q = mL_f  = (252\,\bold{kg})(334,000\,\frac{\bold{J}}{\bold{kg}})=8,4168,000 \,\bold{J} \end{equation*}

So far we need 10,080,000 J + 8,4168,000 J = 94,248,000 J to get ourselves enough water at 0 °C.

Typically the snow is brought to a boil to prevent sickness due to contamination by previous expeditions, and because a hot drink is great for mind and body after hauling a heavy pack and sled through deep snow for 12 hours in sub-zero temperatures at high altitude.

At altitude, the water will boil before reaching 100 °C. According to boiling point vs. altitude graph the boiling point of water at a mountain height of 13,000 ft will be about 190 °F   (88 °C) rather than 212 °F (100°C).  We will only need to raise the water temperature from 0 °C to 88 °C to achieve boiling. Water has a of 4186 J/(kg C°), so we have:

    \begin{equation*} Q = mc\Delta T = (252\, \bold{kg})(4186\,\bold{\frac{J}{kg\, C^{\circ}}})(88\,\frac{\bold{C^{\circ}}}) = 92,828,736 \,\bold{J} \end{equation*}

All told, we need that 92,828,736 plus our previous 94,248,000 J for a total of 187,076,736 J.

According to data on energy densities in fuel, typical liquid fuels (white gas, gasoline, etc.) will provide roughly 40 MJ (40,000,000 J) of net heating energy per kg of fuel burned.  To make our rough estimate the minimum fuel requirement we will assume all of the thermal energy released by burning the fuel is transferred to the water:

    \begin{equation*} Fuel\, Mass = \frac{187,076,736\,\bold{J}}{40,000,000 \,\bold{J/kg}} = 4.7\,\bold{kg} \end{equation*}

The of liquid fuels is less than water at roughly 0.75 kg/L, so this minimum 4.7 kg of fuel comes out to about 6 liters, or 1.5 gallons. Accounting for heat loss from stove and pot to the environment (especially in windy conditions) the actual requirement could end up substantially greater, depending on conditions. On an actual 3-person, 21-day expedition to the summit of Denali in May of 2012 we used just under 2 gallons of fuel, so our estimate was pretty reasonable.

Reinforcement Exercises

Check out the following simulations allows you to play with phase changes.

Energy Forms and Changes

States of Matter



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