Heat Capacities

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 15:04

When we supply heatEnergy transferred as a result of a temperature difference; a form of energy stored in the movement of atomic-sized particles. energyA system's capacity to do work. from a bunsen burner or an electrical heating coil to an object, a rise in temperatureA physical property that indicates whether one object can transfer thermal energy to another object. usually occurs. Provided that no chemical changes or phase changes take place, the rise in temperature is proportional to the quantity of heat energy supplied. If q is the quantity of heat supplied and the temperature rises from T1 to T2 then

q = C × (T2T1)      (1)


q = C × (ΔT)      (1b)

where the constant of proportionality C is called the heat capacityThe quantity of heat energy transfer needed to raise the temperature of a sample of matter by one unit. of the sample. The sign of q in this case is + because the sample has absorbed heat (the change was endothermicIn chemical thermodynamics, describes a process in which energy is transferred from the surroundings to the system as a result of a temperature difference.), and (ΔT) is defined in the conventional way.

If we add heat to any homogenous sample of matterAnything that occupies space and has mass; contrasted with energy. of variable massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object., such as a pure substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. or a solutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture., the quantity of heat needed to raise its temperature is proportional to the mass as well as to the rise in temperature. That is,

q = C × m × (T2T1)      (2)


q = C × m × (Δ T)      (2b)

The new proportionality constant C is the heat capacity per unitA particular measure of a physical quantity that is used to express the magnitude of the physical quantity; for example, the meter is the unit of the physical quantity, length. mass. It is called the specific heat capacityThe quantity of heat energy transfer into a system required to raise the temperature of one gram of a substance by 1 K ( 1 °C). More correctly called specific heat capacity. (or sometimes the specific heatThe quantity of heat energy transfer into a system required to raise the temperature of one gram of a substance by 1 K ( 1 °C). More correctly called specific heat capacity.), where the word specific means “per unit mass.”

Specific heat capacities provide a convenient way of determining the heat added to, or removed from, material by measuring its mass and temperature change. As mentioned [|previously], James JouleThe SI unit of energy (or heat or work); equal to 0.239 calories, or a kg m2/s2 established the connection between heat energy and the intensive propertyA property for which the value does not depend on the quantity of matter under consideration. Density is an example of an intensive property; mass and volume are not intensive properties. temperature, by measuring the temperature change in water caused by the energy released by a falling mass. In an ideal experiment, a 1.00 kg mass falling 10.0 m would release 98.0 J of energy. If the mass drove a propeller immersed in 0.100 literA unit of volume equal to a cubic decimeter. (100 g) of water in an insulated container, its temperature would rise by 0.234oC. This allows us to calculate the specific heat capacity of water:

98 J = C × 100 g × 0.234 oC
C = 4.184 J/goC

At 15°C, the precise value for the specific heat of water is 4.184 J K–1 g–1, and at other temperatures it varies from 4.178 to 4.219 J K–1 g–1. Note that the specific heat has units of g (not the baseIn Arrhenius theory, a substance that increases the concentration of hydroxide ions in an aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) acceptor. In Lewis theory, a species that donates a pair of electrons to form a covalent bond. unit kg), and that since the Centigrade and kelvinA unit of temperature equal to 1/273.16 of the thermodynamic temperature of the triple point of water; the kelvin is the same size as the degree Celsius. The thermodynamic temperature scale (Kelvin scale) has absolute zero as its zero point. scales have identical graduations, either oC or K may be used.

Example 1: How much heat is required to raise the temperature of 500 mL of water (D = 1.0) from 25.0 oC to 75.0 oC, given that the specific heat capacity of water is 4.184 J K–1 g–1?


q = 4.18 J/goC × 500 g × (75.0 - 25.0)
q = 104 500 J or 104 kJ.

Specific heat capacities (25 °C unless otherwise noted)
Substance phase Cp(see below)
air, (Sea level, dry, 0 °C) gas 1.0035
argon gas 0.5203
carbon dioxide gas 0.839
helium gas 5.19
hydrogen gas 14.30
methane gas 2.191
neon gas 1.0301
oxygen gas 0.918
water at 100 °C (steam) gas 2.080
water at T=[1] liquid 0.01°C    4.210
15°C   4.184
25°C    4.181
35°C    4.178
45°C    4.181
55°C    4.183
65°C    4.188
75°C   4.194
85°C    4.283
100°C    4.219
water (ice) at T= [2] solid 0°C      2.050
-10°C    2.0
-20°C    1.943
-40°C    1.818
ethanol liquid 2.44
copper solid 0.385
gold solid 0.129
iron solid 0.450
lead solid 0.127

Electrical Energy Conversion

The most convenient way to supply a known quantity of heat energy to a sample is to use an electrical coil. The heat supplied is the productA substance produced by a chemical reaction. of the applied potential V, the current I flowing through the coil, and the time t during which the current flows:

q = V × I × t      (2)

If the SI unitsThe international system of units (Système International d'Unité) based on seven fundamental units: meter, kilogram, second, ampere, kelvin, candela, mole. volt for applied potential, ampere for current, and second time are used, the energy is obtained in joules. This is because the volt is defined as one joule per ampere per second:

1 volt × 1 ampere × 1 second = 1\begin{matrix}\frac{\text{J}}{\text{A s}}\end{matrix} × 1 A × 1 s = 1 J

EXAMPLE 2: An electrical heating coil, 230 cm3 of water, and a thermometer are all placed in a polystyrene coffee cup. A potential difference of 6.23 V is applied to the coil, producing a current of 0.482 A which is allowed to pass for 483 s. If the temperature rises by 1.53 K, find the heat capacity of the contents of the coffee cup. Assume that the polystyrene cup is such a good insulator that no heat energy is lost from it.

Solution The heat energy supplied by the heating coil is given by

q = V × I × t = 6.23 V × 0.482 A × 483 s = 1450 V A s = 1450 J


q = C × (T2T1)

Since the temperatue rises, T2 > T1 and the temperature change ΔT is positive:

1450 J = C × 1.53 K

so that

\begin{matrix}C=\frac{\text{1450 J}}{\text{1}\text{.53 K}}=\text{948 J K}^{-\text{1}}\end{matrix}

Note: The heat capacity found applies to the complete contents of the cup-water, coil, and thermometer taken together, not just the water.

As discussed in other sections, an older, non-SI energy unit, the calorieA unit of energy equal to 4.184 J; abbreviated cal., was defined as the heat energy required to raise the temperature of 1 g H2O from 14.5 to 15.5°C. Thus at 15°C the specific heat capacity of water is 1.00 cal K–1 g–1. This value is accurate to three significant figures between about 4 and 90°C.

If the sample of matter we are heating is a pure substance, then the quantity of heat needed to raise its temperature is proportional to the amount of substance. The heat capacity per unit amount of substance is called the molar heat capacity, symbol Cm. Thus the quantity of heat needed to raise the temperature of an amount of substance n from T1 to T2 is given by

q = C × n × (T2T1)      (4)

The molar heat capacity is usually given a subscript to indicate whether the substance has been heated at constant pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container. (Cp)or in a closed container at constant volume (CV).

EXAMPLE 3: A sample of neon gas (0.854 mol) is heated in a closed container by means of an electrical heating coil. A potential of 5.26 V was applied to the coil causing a current of 0.336 A to pass for 30.0 s. The temperature of the gas was found to rise by 4.98 K. Find the molar heat capacity of the neon gas, assuming no heat losses.

Solution The heat supplied by the heating coil is given by

   q = V × I × t
= 5.26 V × 0.336 A × 30.0 s
= 53.0 V A s
= 53.0 J

Rearranging Eq. (4), we then have

\begin{matrix}C_{m}=\frac{q}{n\text{(T}_{\text{2}}-\text{T}_{\text{1}}\text{)}}=\frac{\text{53}\text{.0 J}}{\text{0}\text{.854 mol }\times \text{ 4}\text{.98 K}}=\text{12}\text{.47 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\end{matrix}

However, since the process occurs at constant volume, we should write

CV = 12.47 J K–1 mol–1


  1. http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
  2. http://www.engineeringtoolbox.com/ice-thermal-properties-d_576.html