Thermal Mass for Heat Storage - ChemPRIME

Thermal Mass for Heat Storage

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Trombe Walls and Thermal Mass

Many very energy-efficient or "passive houses" use "passive solar" energy storage of various kinds. The simplest is probably the "Trombe Wall". The Trombe wall absorbs and releases large amounts of heat without changing temperature very much, so it must have a high thermal mass or heat capacity.

One Wikipedia article states that if a tank of water were used for the Trombe wall instead of concrete, it could store five times as much heat. Given that rock would be so much heavier, is that possible? Like every solar house designer, we can answer that question with some simple calculations.

Heat Capacities

When the sun supplies heat energy to the Trombe wall, a rise in temperature occurs. In this case, no chemical changes or phase changes take place, so 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 capacity of the wall. The sign of q in this case is + because the sample has absorbed heat (the change was endothermic), and (ΔT) is defined in the conventional way.

If we are interested in comparing Trombe walls of variable mass, the quantity of heat needed to raise the 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 unit mass. It is called the specific heat capacity (or sometimes the specific heat), 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 Joule established the connection between heat energy and the intensive property 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 liter (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 4.218 J K–1 g–1. Note that the specific heat has units of g (not the base unit kg), and that since the Centigrade and kelvin scales have identical graduations, either oC or K may be used.

Example 1: If the sun raises the temperature of a 3 m x 6 m x 0.5 m Trombe wall filled with water (D = 1.0) from 25.0 oC to 35.0 oC, how much heat energy is stored, given that the specific heat capacity of water is 4.184 J K–1 g–1?

Solution: V = 3 m x 6 m x 0.5 m = 9 m3 9 m3 x (100 cm / m)3 = 9 x 106 m3 or 9 x 106 g

q = 4.18 J/goC × 9 x 106 g × (35.0 - 25.0)
q = 3.76 x 108 J or 3.76 x 105 kJ.

Example 2: If the sun raises the temperature of a 3 m x 6 m x 0.5 m Trombe wall made of concrete (typical D = 2.3 g/cm3) from 25.0 oC to 35.0 oC, how much heat energy is stored, given that the specific heat capacity of concrete (see below) is 0.880 J K–1 g–1?

Solution: V = 3 m x 6 m x 0.5 m = 9 m3 9 m3 x (100 cm / m)3 = 9 x 106 m3 m = D x V = 2.3 g/cm3 x 9 x 106 m3 = 2.07 x 107 g

q = 0.880 J/goC × 2.07 x 107 g × (35.0 - 25.0)
q = 1.82 x 108 J or 1.8 x 105 kJ.

Note that the water can absorb about 2 times as much heat for the same volume and same temperature change. For the same mass, however, water can absorb 4.18/0.880 = 4.75 times as much heat. The mass-based calculation must be the basis for the Wikipedia claim.

Specific heat capacity of building materials

(Usually of interest to builders and solar designers)

Specific heat capacity of building materials [4]
Substance Phase cp
Asphalt solid 0.920
Brick solid 0.840
Concrete solid 0.880
Glass, silica solid 0.840
Glass, crown solid 0.670
Glass, flint solid 0.503
Glass, pyrex solid 0.753
Granite solid 0.790
Gypsum solid 1.090
Marble, mica solid 0.880
Sand solid 0.835
Soil solid 0.800
Wood solid 0.420
Substance Phase cp

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 100 °C liquid 4.184
ethanol liquid 2.44
water at -10 °C (ice)) solid 2.05
copper solid 0.385
gold solid 0.129
iron solid 0.450
lead solid 0.127

Other Heat Storage Strategies

Molten salt can be used to store energy at a higher temperature, so that stored solar energy can be used to boil water to run steam turbines. The sodium nitrate/potassium nitrate salt mixture melts at 221 °C (430 °F). It is kept liquid at 288 °C (550 °F) in an insulated "cold" storage tank. The liquid salt is pumped through panels in a solar collector where the focused sun heats it to 566 °C (1,051 °F). It is then sent to a hot storage tank. This is so well insulated that the thermal energy can be usefully stored for up to a week.

When electricity is needed, the hot salt is pumped to a conventional steam-generator to produce superheated steam for a turbine/generator as used in any conventional coal, oil or nuclear power plant. A 100-megawatt turbine would need tanks of about 30 feet (9.1 m) tall and 80 feet (24 m) in diameter to drive it for four hours by this design.

To understand energy conversion from thermal to electrical, we need to know something about electrical units.

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 product 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 units 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 3: 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 calorie, 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 pressure (Cp)or in a closed container at constant volume (CV).

EXAMPLE 4: 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


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