Heat capacity is the capability of a product to absorb warm without directly reflecting every one of it together a increase in temperature. You should read the sections on heat and temperature as background, and the water ar would help, too.

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As heat is included uniformly to choose quantities of various substances, their temperatures deserve to rise at different rates. For example, metals,


good conductors that heat, display fast temperature rises once heated. It is fairly easy to warmth a steel until that glows red. On the various other hand, water can absorb a lot of warmth with a relatively small rise in temperature. Insulating products (insulators) are really poor conductors of heat, and are offered to isolate materials that must be kept at various temperatures — favor the within of your residence from the outside.


This graph reflects the climb in temperature as warm is added at the same price to equal masses the aluminium (Al) and water (H2O). The temperature the water rises much more slowly than that the Al.

In the metal, Al atom only have actually translational kinetic power (although that motion is combination strongly come neighbor atoms). Water, ~ above the various other hand, can rotate and vibrate together well. This degrees of freedom of motion can absorb kinetic energy without reflecting it as a rise in temperature of the substance.


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Most substances obey the legislation of equipartition the energy end a broad range of temperatures. The law says that power tends to be spread evenly among all of the levels of freedom of a molecule — translation, rotation and also vibration. This has aftermath for building material with much more or fewer atoms. In the diagram below, each container to represent a degree of freedom. The situations for a 3-atom and a 10-atom


molecule are shown. If the same full amount the heat energy is included to each molecule, the 3-atom molecule ends up with more energy in its translational degrees of freedom. Since the 10-atom molecule has much more vibrational modes in which to keep kinetic energy, much less is accessible to get in the translational modes, and also it is mostly the translational power that we measure as temperature.


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There"s one an ext refinement left to do to heat capacity. Obviously, the lot of heat compelled to advanced the temperature that a huge quantity that a problem is better than the amount compelled for a tiny amount the the same substance.

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


Specific heats of a good many substances have actually been measure under a variety of conditions. They room tabulated in books an on-line.

We generally pick units that J/gram or KJ/Kg. The particular heat of liquid water is 4.184 J/g, i beg your pardon is additionally 4.184 KJ/Kg. The calorie is a unit the heat identified as the amount of heat compelled to raise the temperature that 1 cm3 the water by 1˚C.


Specific heat is the heat capacity per unit mass.

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


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

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

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


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

Using this formula, it"s reasonably easy come calculate warm added, final or initial temperature or the particular heat chin (that"s exactly how it"s measured) if the other variables space known.


The warm q included or advanced for a temperature change of a fixed m that a substance with particular 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 the it"s OK to swap ˚C for K due to the fact that the dimension of the Celsius degree and also the Kelvin are the same.


Example 1

Calculate the amount of warm (in Joules) required to change the temperature of 1 Liter the water (1 together = 1 Kg) indigenous 20˚C come 37˚C.


The specific heat capacity (C) of water is 4.184 J/g˚C (or J/g·K — as long we work with Celsius degrees or Kelvins, the ΔT will certainly be the same because the dimension of the two room 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 massive of 1 together of water (the gram is characterized as the fixed of 1 mL of water), and also the temperature change (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 energy runs end 1,000, we typically express the lot in KiloJoules (KJ) in bespeak to leveling the number.


Practice problems

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

1. How much heat (in Joules) does it require to raise the temperature of 100 g of H2O indigenous 22˚C to 98˚C? Solution
2. If that takes 640 J the heat power to boost the temperature that 100 g of a substance by 25˚C (without changing its phase), calculation the particular heat the the substance. Solution
3. If 80 J of warm are included to 100 ml that ethanol at first at 10˚C, calculate the last temperature that the sample. Solution

× difficulty 1 equipment

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


× problem 2 systems

Rearrange the warmth equation to deal with 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: when calculating Δ T"s, its" yes sir to usage either Celsius degrees or Kelvins, since the size, and therefore any type of difference, will be the same. That all drops apart with Fahrenheit, though.


× trouble 3 solution

First rearrange the warmth equation to resolve 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 number of grams that ethanol making use of the density and also being mindful to follow the units.

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

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 through a phase-change temperature

Phase transforms are a huge source or sink that heat. Here, because that example, is the heating curve of water.

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It shows the rise in temperature as heat is included at a continuous rate to water. Here"s what"s going on in regions A-E:

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


B. Solid ice cream is melted to fluid water. During the enhancement of the latent heat of blend (ΔHf), no temperature increase is observed, however hydrogen bond holding the ice together break.

C. warm is included to fluid water over 0˚C, and also its temperature rises at a continuous rate till the boiling point at 100˚C.

D. Water at 100˚C absorbs a good deal the heat power at 100˚C together it experience a phase shift from liquid to gas. This is the latent warmth of vaporization, ΔHv, the energy it takes because that water to have actually no more cohesive force.

E. Finally, gas water above 100˚C absorbs heat, raising its temperature in ~ a constant rate. Water has no much more phase transitions after ~ this.

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

Enthalpies of blend and vaporization room tabulated and also can be looked up. The Wikipedia web page of a compound is commonly a good place to discover them. Listed below we"ll do an instance of a warmth calculation as the temperature of a substance rises v a step change.


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Cohesive forces

Cohesive pressures are forces that organize a substance together. When water hits a waxy or hydrophobic surface, that forms tiny sphere-like drops – "beads." this beads that water minimization the call with the surface and with the air, and maximize the contact of water with itself. Liquid water is really cohesive. It forms intermittent, but reasonably strong bonds with itself.

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


The heat absorbed or exit upon a phase change is calculated by multiplying the enthalpy of vaporization, ΔHv, or the enthalpy the fusion, ΔHf by the variety of moles that substance:

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

The enthalpy of blend is often dubbed the "latent warmth of fusion" and the enthalpy that vaporization is often referred to as the "latent warmth of vaporization."

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


Solution: there is a phase transition the water in this temperature range, for this reason this difficulty will comprise three steps:

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

Here space the calculations because that each of ours steps:

Step 1:The quantity of heat forced to progressive the temperature of ice (before the melts) by 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 that 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 lot of heat required to advanced the temperature of fluid water through 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 compare this to a similar calculation, however this time we"ll warm liquid water v its boiling allude to a gas.


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

progressive the temperature of fluid water from 80˚C to the boiling point, 100˚C, utilizing the particular heat that water, C = 4.184 J·g-1K-1. Transform water to steam (gaseous water) at 100˚C, utilizing the molar enthalpy that vaporization, ΔHf = 2258 J·g-1. Progressive the temperature of vapor from 100˚C come 125˚C, using the details heat the steam, C = 2.010 J·g-1K-1.

Here are the calculations because that each of our steps:

Step 1:The amount of heat compelled to raise the temperature of water (before the vaporizes) native 80˚C come 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. Below we use the heat of vaporization that water:

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

Step 3: Finally, we calculate the amount of heat forced to change the temperature of the 80 g of steam from 100˚C to 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 action is to add all of these energies together:

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

Notice the the largest contribution come this energy, by far, is in evaporating the water — an altering it from liquid to gas. This process takes a tremendous amount of energy, and that energy accounts for the huge amount of power it takes to cook water to make vapor in electric generating plants of all kinds (including nuclear), and for the efficient way humans have actually of cooling our bodies: perspiration.


Practice problems

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

1. How much warmth (in Joules) go it take to readjust 120g of ice cream at -10˚C to 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 of the water come 37˚C (human human body temp.). Solution
2. How much heat is released as soon as 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 the water in ~ 25˚C to totally melt 50g of ice at 0˚C? (ΔHf = 334 KJ·Kg-1) Solution

First calculation the warmth needed come raise the temperature the water from 10˚C to 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 liquid water in ~ 0˚C:

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

Finally, rais the temperature the the water to 37˚C, and include 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 many of the power goes 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 heavy steam to liquid 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.

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