Heat capacity is the ability of a product to absorb warmth without straight reflecting every one of it together a rise in temperature. You should read the part on heat and temperature together background, and also the water ar would help, too.

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As warm is included uniformly to choose quantities of various substances, your temperatures deserve to rise at various rates. Because that example, metals,


good conductors the heat, display fast temperature rises when heated. That is reasonably easy to warmth a steel until that glows red. ~ above the various other hand, water deserve to absorb a lot of of heat with a reasonably small rise in temperature. Insulating products (insulators) are really poor conductors the heat, and also are offered to isolate materials that have to be kept at various temperatures — like the inside of your house from the outside.


This graph mirrors the increase in temperature as warmth is included at the same rate to same masses the aluminium (Al) and also water (H2O). The temperature of water rises much an ext slowly than that that Al.

In the metal, Al atoms only have actually translational kinetic power (although that motion is coupled strongly come neighbor atoms). Water, top top the other hand, deserve to rotate and vibrate as well. This degrees that freedom the motion deserve to absorb kinetic energy without reflecting it as a increase in temperature of the substance.


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Most substances follow the legislation of equipartition the energy over a broad variety of temperatures. The law says that power tends come be dispersed evenly among every one of the degrees of liberty of a molecule — translation, rotation and vibration. This has results for building materials with more or fewer atoms. In the diagram below, each container represents a degree of freedom. The situations for a 3-atom and also a 10-atom


molecule room shown. If the same complete amount the heat power is added to every molecule, the 3-atom molecule ends up with more energy in its translational degrees of freedom. Because the 10-atom molecule has an ext vibrational modes in i beg your pardon to store kinetic energy, much less is easily accessible to go into the translational modes, and also it is largely the translational energy that we measure as temperature.


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There"s one much more refinement left to do to warm capacity. Obviously, the quantity of heat compelled to raise the temperature of a huge quantity of a substance is higher than the amount required for a tiny amount that the very same substance.

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


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

We generally pick units the J/gram or KJ/Kg. The details heat of fluid water is 4.184 J/g, i m sorry is likewise 4.184 KJ/Kg. The calorie is a unit of heat characterized as the amount of heat compelled to advanced the temperature the 1 cm3 the water by 1˚C.


Specific heat is the warmth capacity every unit mass.

The particular heat of water is 1 cal/g˚C = 4.184J/g˚C


The heat, q, required to raise the temperature of a mass, m, the a substance by an quantity ΔT is

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

where C is the certain heat and Tf and Ti room the final and also initial temperatures.


The steep of a graph the temperature vs. Heat included to a unit mass is simply 1/C.

Using this formula, it"s relatively easy come calculate heat added, last or early stage temperature or the specific heat chin (that"s exactly how it"s measured) if the various other variables space known.


The warm q added or advanced for a temperature change of a massive m the a problem with certain heat C is

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

The systems of details heat room 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 size of the Celsius degree and also the Kelvin space the same.


Example 1

Calculate the amount of heat (in Joules) compelled to change the temperature that 1 Liter the water (1 together = 1 Kg) indigenous 20˚C to 37˚C.


The specific heat capacity (C) of water is 4.184 J/g˚C (or J/g·K — as lengthy we job-related with Celsius levels or Kelvins, the ΔT will be the same since the size 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 because that the mass of 1 l of water (the gram is characterized as the mass of 1 mL that water), and also the temperature adjust (37˚C - 20˚C), we get:


$$= (1000 , g) left( 4.185 fracJgcdot ˚C ight) (37 - 20)˚C$$

The an outcome is

$$= 71,128 ; J = f 71 ; KJ$$

When the number of Joules of power runs over 1,000, we generally express the amount in KiloJoules (KJ) in order to simplify the number.


Practice problems

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

1. How much warm (in Joules) walk it require to raise the temperature of 100 g that H2O indigenous 22˚C come 98˚C? Solution
2. If it takes 640 J of heat energy to boost the temperature the 100 g the a substance by 25˚C (without transforming its phase), calculate the details heat of the substance. Solution
3. If 80 J of warm are added to 100 ml of ethanol initially at 10˚C, calculate the final temperature of the sample. Solution

× problem 1 systems

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


× problem 2 equipment

Rearrange the warmth equation to settle for C:

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

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

Note: once calculating Δ T"s, its" yes sir to usage either Celsius degrees or Kelvins, because the size, and therefore any difference, will certainly be the same. The all falls apart with Fahrenheit, though.


× problem 3 equipment

First rearrange the warmth equation to solve for the last temperature.

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

Now calculate the variety of grams the ethanol making use of the density and also being mindful to follow the units.

$$ eginalign 100 & imes 10^3 , l left( frac1 , m^310^3 , L ight) left( frac785 , Kgm^3 ight) left( frac1000 , g1 , Kg ight)\ &= 78.9 ext g that ethanol endalign$$

Now finish:

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


Heat (enthalpy) of step change

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

Phase alters are a large source or sink that heat. Here, for example, is the heater curve that water.

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It reflects the increase in temperature as warmth is added at a consistent rate to water. Here"s what"s going on in regions A-E:

A. warmth is included to hard water (ice) below 0˚C, and also its temperature rises at a constant rate.


B. Solid ice is melted to liquid water. During the addition of the latent warm of blend (ΔHf), no temperature increase is observed, however hydrogen bonds holding the ice with each other break.

C. warmth is included to fluid water over 0˚C, and its temperature rises in ~ a continuous rate until the boiling suggest at 100˚C.

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

E. Finally, gaseous water over 100˚C absorbs heat, raising its temperature at a consistent rate. Water has actually no an ext phase transitions after ~ this.

The relatively huge attractive intermolecular forces in between water molecules gives water really high heats of blend and vaporization. Compared to most other substances, that takes a large amount of warmth to melt water ice and also to cook or evaporate water.

Enthalpies of combination and vaporization room tabulated and also can be looked up. The Wikipedia page of a compound is usually a an excellent place to find them. Listed below we"ll do an example of a heat calculation as the temperature the a substance rises through a phase change.


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Cohesive forces

Cohesive forces are forces that hold a problem together. As soon as water hits a waxy or hydrophobic surface, it forms small sphere-like fall – "beads." these beads that water minimization the contact with the surface and also with the air, and maximize the contact of water through itself. Fluid water is really cohesive. It creates intermittent, but fairly strong bonds v itself.

Other substances like CO2 lack such solid intermolecular attractions, and don"t type liquids or solids unless really cold or at an extremely high pressure.


The heat took in or exit upon a phase shift is calculation by multiplying the enthalpy of vaporization, ΔHv, or the enthalpy the fusion, ΔHf through the variety of moles the substance:

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

The enthalpy of blend is often called the "latent heat of fusion" and the enthalpy that vaporization is often dubbed the "latent heat of vaporization."

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


Solution: over there is a phase transition the water in this temperature range, so this problem will consist of three steps:

progressive the temperature of ice from -20˚C come the melting point, 0˚C, utilizing the details heat the ice, C = 2.010 J·g-1K-1. Transform ice to water at 0˚C, utilizing the molar enthalpy the fusion, ΔHf = 333.5 J·g-1. Raise the temperature of fluid water indigenous 0˚C come 25˚C, using the specific heat of water, C = 4.184 J·g-1K-1.

Here space the calculations for each of ours steps:

Step 1:The amount of heat required to advanced the temperature of ice (before the melts) through 20˚C is:

$$ eginalign q &= m C Delta T \ &= (18 , g) left( 2.01 fracJgcdot K ight)(273 , K - 263 , K) \ &= f 723.6 ; J endalign$$


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

Step 2: The amount of heat required to melt 18 g of ice cream is:

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

Step 3: The quantity of heat compelled to advanced the temperature of liquid water by 25˚C is:

$$ eginalign q &= m C Delta T \ &= (18 , g) left( 4.184 fracJgcdot K ight)(298 , K - 273 , K) \ &= f 1882.8 ; J endalign$$

Adding every one of these energies up, we acquire the total, q = 2642 J,

Now let"s to compare this to a comparable calculation, however this time we"ll warm liquid water with its boiling suggest to a gas.


Solution: This is additionally a three-step problem, yet this time we"re vaproizing water. Here are the steps:

advanced the temperature of fluid water from 80˚C to the cook point, 100˚C, using the details heat the water, C = 4.184 J·g-1K-1. Transform water to steam (gaseous water) in ~ 100˚C, making use of the molar enthalpy of vaporization, ΔHf = 2258 J·g-1. Advanced the temperature of vapor from 100˚C come 125˚C, utilizing the specific heat the steam, C = 2.010 J·g-1K-1.

Here space the calculations because that each of our steps:

Step 1:The quantity of heat required to advanced the temperature the water (before that vaporizes) from 80˚C to 100˚C is:

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


Step 2: convert the fluid water to vapor at 100˚C. Here we usage the heat of vaporization the water:

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

Step 3: Finally, we calculate the lot of heat forced to change the temperature of the 80 g of steam from 100˚C come 125˚C:

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

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

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

Notice that the biggest contribution to this energy, by far, is in evaporating the water — an altering it from liquid to gas. This procedure takes a remarkable amount the energy, and also that power accounts for the large amount of energy it take away to cook water come make vapor in electric generating tree of all kinds (including nuclear), and also for the efficient method humans have actually of cooling our bodies: perspiration.


Practice problems

(Use the table below to look increase missing details heats; heats of combination or vaporization are offered in the problems.)

1. How much warm (in Joules) go it take it to change 120g of ice cream at -10˚C come water in ~ 37˚C? (ΔHf = 334 KJ·Kg-1)? note that this is a three-step problem: an initial heat the ice cream to 0˚C, then convert all 120g come liquid, then raise the temperature the the water to 37˚C (human body temp.). Solution
2. How much warm is released when 1 Kg of vapor 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 at 0˚C? (ΔHf = 334 KJ·Kg-1) Solution

First calculation the warmth needed to raise the temperature of water native 10˚C come 0˚C

$$ eginalign q &= mcDelta T \ &= (120 , g) left( 2.11 fracJg˚C ight)(0 - (-10))˚C \ &= 2532 ; J endalign$$

Now convert 120 g of ice cream at 0˚C to fluid water at 0˚C:

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

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

$$ eginalign q &= mcDelta T \ &= (120 , g) left( 4.184 fracJg˚C ight)(37 - 0)˚C \ &= 18,577 ; J endalign$$

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

Notice that most of the energy goes right into breaking the continual crystal lattice framework of the ice (melting it).


Cool the vapor from 300˚C to 100˚C.

$$ eginalign q &= mcDelta T \ &= (1000 , g) left( 2.08 fracJg˚C ight)(-200˚C) \ &= -416,000 ; J endalign$$

Convert the vapor to fluid at 100˚C.

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

Cool the fluid from 100˚C come 40˚C.

$$ eginalign q &= mcDelta T \ &= (1000 , g) left( 4.184 fracJg˚C ight)(-60˚C) \ &= -251,040 ; J endalign$$