Heat capacity is the capacity of a product to absorb warm without directly reflecting all of it as a increase in temperature. You have to read the part on heat and temperature together background, and also the water section would help, too.

You are watching: Heat of vaporization of water kj kg

As heat is included uniformly to favor quantities of various substances, your temperatures can rise at different rates. Because that example, metals,

good conductors of heat, display fast temperature rises as soon as heated. That is fairly easy to heat a steel until it glows red. ~ above the other hand, water deserve to absorb a many of warmth with a fairly small rise in temperature. Insulating materials (insulators) are really poor conductors that heat, and also are offered to isolate materials that must be kept at various temperatures — prefer the within of your residence from the outside.

This graph mirrors the rise in temperature as warm is included at the same price to same masses that aluminium (Al) and water (H2O). The temperature of water rises much much more slowly 보다 that of Al.

In the metal, Al atoms only have actually translational kinetic power (although that activity is coupled strongly to neighbor atoms). Water, on the various other hand, have the right to rotate and also vibrate as well. These degrees the freedom the motion deserve to absorb kinetic energy without mirroring it together a climb in temperature that the substance.


Most substances follow the regulation of equipartition that energy over a broad selection of temperatures. The legislation says that power tends come be spread evenly among all of the levels of freedom of a molecule — translation, rotation and vibration. This has consequences for building material with much more or under atoms. In the diagram below, every container represents a degree of freedom. The instances for a 3-atom and a 10-atom

molecule are shown. If the same total amount the heat power is added to each molecule, the 3-atom molecule ends up with more energy in that translational levels of freedom. Since the 10-atom molecule has an ext vibrational modes in which to store kinetic energy, less is available to enter the translational modes, and also it is mainly the translational power that we measure together temperature.


There"s one an ext refinement left to do to warm capacity. Obviously, the amount of heat compelled to advanced the temperature of a large quantity of a problem is higher than the amount required for a small amount that the same substance.

To manage for the amount, we typically measure and also report warmth capacities together specific heat, the warm capacity per unit mass.

Specific heats of a great many substances have been measured under a selection of conditions. They are tabulated in publications an on-line.

We generally pick units of J/gram or KJ/Kg. The specific heat of fluid water is 4.184 J/g, i beg your pardon is also 4.184 KJ/Kg. The calorie is a unit that heat characterized as the quantity of heat forced to raise the temperature of 1 cm3 that water by 1˚C.

Specific heat is the warmth capacity per unit mass.

The specific heat that water is 1 cal/g˚C = 4.184J/g˚C

The heat, q, compelled to advanced the temperature the a mass, m, that a problem by an amount ΔT is

$$q = mC \Delta T = mC (T_f - T_i)$$

where C is the specific heat and Tf and Ti are the final and also initial temperatures.

The slope of a graph of temperature vs. Heat included to a unit massive is simply 1/C.

Using this formula, it"s fairly easy to calculate warmth added, final or initial temperature or the particular heat chin (that"s exactly how it"s measured) if the other variables room known.

The warmth q included or advanced for a temperature change of a mass m that a substance with certain heat C is

$$q = mC \Delta T = mC (T_f - T_i)$$

The devices of particular heat space usually J/mol·K (J·mol-1K-1) or J/g·K (J·g-1·K-1). Remember that it"s OK come swap ˚C because that K since the dimension of the Celsius degree and the Kelvin are the same.

Example 1

Calculate the lot of heat (in Joules) forced to readjust the temperature of 1 Liter of water (1 l = 1 Kg) indigenous 20˚C to 37˚C.

The specific heat volume (C) the water is 4.184 J/g˚C (or J/g·K — as long we occupational with Celsius degrees or Kelvins, the ΔT will be the same due to the fact that the dimension of the two are the same. It"s Fahrenheit that"s a smaller-sized degree). The equation we need is:

$$q = m C \Delta T$$

Plugging in 1000 g for the mass of 1 l of water (the gram is identified as the fixed of 1 mL of water), and the temperature adjust (37˚C - 20˚C), we get:

$$= (1000 \, g) \left( 4.185 \fracJg\cdot ˚C \right) (37 - 20)˚C$$

The result is

$$= 71,128 \; J = \bf 71 \; KJ$$

When the variety of Joules of energy runs end 1,000, we usually express the lot in KiloJoules (KJ) in stimulate to simplify the number.

Practice problems

(Use the table listed below to look up missing certain heats.)

1. How much warm (in Joules) go it take to raise the temperature of 100 g that H2O from 22˚C to 98˚C? Solution
2. If the takes 640 J that heat power to increase the temperature of 100 g the a substance by 25˚C (without an altering its phase), calculation the details heat that the substance. Solution
3. If 80 J of warmth are added to 100 ml of ethanol originally at 10˚C, calculation the final temperature that the sample. Solution

× difficulty 1 systems

$$ \beginalign q &= mC \Delta T \\ &= mC (T_f - T_i) \\ \\ &= (100 \, g) \left( 4.184 \fracJg˚C \right) (98 - 22)˚C \\ \\ &= 31,789 \; J \\ &= 31.8 \; KJ \endalign$$

× difficulty 2 solution

Rearrange the warmth equation to resolve for C:

$$q = mC \Delta T \; \longrightarrow \; C = \fracqm \Delta T$$

$$ \beginalign C &= \fracqm \Delta T \\ \\ &= \frac640 \; J(100 \, g)(25˚C) \\ \\ &= 0.256 \fracJg ˚C \endalign$$

Note: as soon as calculating Δ T"s, its" yes sir to usage either Celsius levels or Kelvins, since the size, and therefore any difference, will be the same. It all drops apart v Fahrenheit, though.

× trouble 3 solution

First rearrange the warmth equation to deal with for the final temperature.

$$ \beginalign q = mC \Delta T \; &\longrightarrow \; T_f - T_i = \fracqmC \\ &\longrightarrow T_f = \fracqmC + T_i \endalign$$

Now calculation the variety of grams that ethanol using the density and also being cautious to monitor the units.

$$ \beginalign 100 &\times 10^3 \, l \left( \frac1 \, m^310^3 \, L \right) \left( \frac785 \, Kgm^3 \right) \left( \frac1000 \, g1 \, Kg \right)\\ &= 78.9 \text g that ethanol \endalign$$

Now finish:

$$ \beginalign T_f &= \fracqmC + T_i \\ &= \frac80 \, J(78.9 \, g)\left( 2.44 \fracJg˚C \right) + 10˚C \\ &= 10.42˚C \endalign$$

Heat (enthalpy) of step change

... Or, what if we warmth or cool v a phase-change temperature

Phase transforms are a large source or sink of heat. Here, for example, is the heating curve of water.


It shows the increase in temperature as warmth is included at a constant rate to water. Here"s what"s walk on in regions A-E:

A. heat is added to heavy water (ice) below 0˚C, and also its temperature rises in ~ a constant rate.

B. Solid ice is melted to fluid water. Throughout the addition of the latent warm of combination (ΔHf), no temperature climb is observed, yet hydrogen binding holding the ice with each other break.

C. heat is added to fluid water over 0˚C, and its temperature rises at a constant rate till the boiling suggest at 100˚C.

D. Water at 100˚C absorbs a good deal of heat power at 100˚C as it undergoes a phase shift from fluid to gas. This is the latent warm of vaporization, ΔHv, the energy it takes because that water to have no much more cohesive force.

E. Finally, gas water above 100˚C absorbs heat, increasing its temperature in ~ a constant rate. Water has no an ext phase transitions ~ this.

The relatively big attractive intermolecular forces in between water molecules gives water an extremely high heats of fusion and vaporization. Compared to many other substances, it takes a large amount of warm to melt water ice and to cook or evaporate water.

Enthalpies of blend and vaporization are tabulated and also can it is in looked up. The Wikipedia page of a compound is generally a great place to find them. Listed below we"ll do an example of a heat calculation together the temperature the a problem rises with a step change.


Cohesive forces

Cohesive forces are pressures that organize a substance together. When water access time a waxy or hydrophobic surface, that forms tiny sphere-like autumn – "beads." these beads the water minimize the contact with the surface and with the air, and maximize the call of water with itself. Liquid water is really cohesive. It forms intermittent, but fairly strong bonds through itself.

Other substances like CO2 absence such strong intermolecular attractions, and also don"t type liquids or solids unless very cold or at really high pressure.

The heat absorbed or released upon a phase transition is calculate by multiplying the enthalpy of vaporization, ΔHv, or the enthalpy that fusion, ΔHf by the variety of moles that substance:

$$ \beginalign q &= m \, \Delta H_f \\<5pt> q &= m \, \Delta H_v \endalign$$

The enthalpy of blend is often called the "latent warmth of fusion" and the enthalpy that vaporization is often called the "latent warm of vaporization."

The devices of ΔHf and also ΔHv room Joules/mole (J·mol-1) or J/g (J·g-1).

Solution: there is a phase transition that water in this temperature range, so this problem will make up three steps:

advanced the temperature of ice cream from -20˚C to the melting point, 0˚C, making use of the details heat of ice, C = 2.010 J·g-1K-1. Transform ice to water at 0˚C, making use of the molar enthalpy that fusion, ΔHf = 333.5 J·g-1. Progressive the temperature of fluid water native 0˚C to 25˚C, utilizing the certain heat that water, C = 4.184 J·g-1K-1.

Here space the calculations because that each of our steps:

Step 1:The amount of heat forced to progressive the temperature of ice cream (before that melts) by 20˚C is:

$$ \beginalign q &= m C \Delta T \\ &= (18 \, g) \left( 2.01 \fracJg\cdot K \right)(273 \, K - 263 \, K) \\ &= \bf 723.6 \; J \endalign$$

Note that we"ve convert the Celsius temperatures to Kelvin.

Step 2: The quantity of heat compelled to melt 18 g of ice is:

$$ \beginalign q &= m \Delta H_f \\ &= (18 \, g) \left( 2.01 \fracJg \right) \\ &= 36.18 \; J \endalign$$

Step 3: The lot of heat forced to raise the temperature of fluid water through 25˚C is:

$$ \beginalign q &= m C \Delta T \\ &= (18 \, g) \left( 4.184 \fracJg\cdot K \right)(298 \, K - 273 \, K) \\ &= \bf 1882.8 \; J \endalign$$

Adding all of these energies up, we get the total, q = 2642 J,

Now let"s compare this come a similar calculation, but this time we"ll warmth liquid water through its boiling allude to a gas.

Solution: This is likewise a three-step problem, but this time we"re vaproizing water. Right here are the steps:

progressive the temperature of liquid water indigenous 80˚C come the cook point, 100˚C, making use of the particular heat of water, C = 4.184 J·g-1K-1. Convert water to vapor (gaseous water) at 100˚C, using the molar enthalpy of vaporization, ΔHf = 2258 J·g-1. Progressive the temperature of steam from 100˚C to 125˚C, making use of the specific heat of steam, C = 2.010 J·g-1K-1.

Here room the calculations for each of ours steps:

Step 1:The quantity of heat forced to raise the temperature of water (before the vaporizes) native 80˚C come 100˚C is:

$$ \beginalign q &= m C \Delta T \\ &= (18 \, g)\left(4.184 \fracJmol \, ˚C\right)(100 - 80)˚C \\ &= 1505 \; J \endalign$$

Step 2: transform the liquid water to vapor at 100˚C. Here we usage the warm of vaporization the water:

$$ \beginalign q &= m \Delta H_v = (18 \, g)\left( 2258 \, \fracJg \right) \\ &= 180,640 \; J \endalign$$

Step 3: Finally, us calculate the lot of heat required to adjust the temperature that the 80 g of steam from 100˚C come 125˚C:

$$ \beginalign q &= m C \Delta T \\ &= (18 \, g)\left( 2.010 \, \fracJmol ˚C \right)(125 - 100)˚C \\ &= 904 \; J \endalign$$

Almost there. The last step is to include all of these energies together:

$$ \beginalign q_total &= 1505 \, J + 180,640 \, J + 904 \, J \\ &= 183,050 \; J = 183 \; KJ \endalign$$

Notice the the biggest contribution come this energy, by far, is in evaporating the water — changing it from liquid to gas. This procedure takes a significant amount the energy, and that power accounts because that the large amount of power it takes to boil water to make vapor in electrical generating tree of every kinds (including nuclear), and for the efficient way humans have actually of cooling our bodies: perspiration.

Practice problems

(Use the table below to look up missing specific heats; heats of blend or vaporization are provided in the problems.)

1. How much warmth (in Joules) walk it take to adjust 120g of ice cream at -10˚C come water at 37˚C? (ΔHf = 334 KJ·Kg-1)? note that this is a three-step problem: first heat the ice to 0˚C, then convert all 120g to liquid, climate raise the temperature that the water come 37˚C (human human body temp.). Solution
2. How much heat is released once 1 Kg of heavy steam at 300˚C is cooled to fluid at 40˚C? (ΔHv = 2260 KJ·Kg-1) Solution
3. Is there enough heat in 100 ml that water in ~ 25˚C to fully melt 50g of ice cream at 0˚C? (ΔHf = 334 KJ·Kg-1) Solution

First calculate the heat needed to raise the temperature the water from 10˚C come 0˚C

$$ \beginalign q &= mc\Delta T \\ &= (120 \, g) \left( 2.11 \fracJg˚C \right)(0 - (-10))˚C \\ &= 2532 \; J \endalign$$

Now transform 120 g of ice at 0˚C to fluid water in ~ 0˚C:

$$ \beginalign q &= m \Delta H_f \\ &= (120 \, g)(334 \, J/g) \\ &= 40,080 \; J \endalign$$

Finally, rais the temperature of the water come 37˚C, and include up the energies:

$$ \beginalign q &= mc\Delta T \\ &= (120 \, g) \left( 4.184 \fracJg˚C \right)(37 - 0)˚C \\ &= 18,577 \; J \endalign$$

$$q_total = 2352 \, J + 40080 \, J _ 18577 \, J = 61 \; KJ$$

Notice that most of the power goes into breaking the constant crystal lattice framework of the ice cream (melting it).

Cool the heavy steam from 300˚C to 100˚C.

$$ \beginalign q &= mc\Delta T \\ &= (1000 \, g) \left( 2.08 \fracJg˚C \right)(-200˚C) \\ &= -416,000 \; J \endalign$$

Convert the heavy steam to liquid at 100˚C.

See more: Love Is A Many Splendored Thing Meaning, Love Is A Many Splendored Thing

$$ \beginalign q &= -m \Delta H_f \\ &= -(1000 \, g)(2260 \, J/g) \\ &= -2,260,000 \; J \endalign$$

Cool the liquid from 100˚C to 40˚C.

$$ \beginalign q &= mc\Delta T \\ &= (1000 \, g) \left( 4.184 \fracJg˚C \right)(-60˚C) \\ &= -251,040 \; J \endalign$$