Specific heat (C e ) is the amount of heat that must be applied to a unit mass of a material in order to raise its temperature by one unit . It is an intensive thermal property of matter, that is, it does not depend on the extent of the material or its quantity, but only on its composition. In this sense, it is a characteristic property that is of great importance to determine the possible applications of each material, and that helps to determine part of the thermal behavior of substances when they come into contact with bodies or media that are at different temperatures.
From a certain point of view we could say that specific heat corresponds to the intensive version of heat capacity (C), defining it as the amount of heat that must be supplied to a system to increase its temperature by one unit. It can also be understood as the constant of proportionality between the heat capacity of a system (a body, a substance, etc.) and its mass.
The value of the specific heat of a substance depends on whether the heating (or cooling) is carried out at constant pressure or at constant volume. This gives rise to two specific heats for each substance, namely the specific heat at constant pressure (C P ) and the specific heat at constant volume (C V ). However, the difference can only be seen in gases, so for liquids and solids we usually just talk about dry specific heat.
specific heat formula
We know from experience that the heat capacity of a body is proportional to its mass, that is, that
As we mentioned in the previous section, the specific heat represents the proportionality constant between these two variables, so the above proportionality relationship can be written in the form of the following equation:
We can solve this equation to get an expression for the specific heat:
On the other hand, we know that the heat capacity is the constant of proportionality between the heat (q) that is required to increase the temperature of a system by an amount ΔT and said increase in temperature. In other words, we know that q = C * ΔT. Combining this equation with the heat capacity equation shown above, we get:
Solving this equation to find the specific heat, we obtain a second equation for it:
Specific Heat Units
The last equation obtained for specific heat shows that the units of this variable are [q][m] -1 [ΔT] -1 , that is, heat units over mass and temperature units. Depending on the system of units in which you are working, these units can be:
Unit system Specific heat units International system J.kg -1 .K -1 which is equivalent to am 2 ⋅K − 1 ⋅s − 2 imperial system BTU⋅lb − 1 ⋅°F − 1 calories cal.g -1 .°C -1 which is equivalent to Cal.kg -1 .°C -1 other units kJ.kg -1 .K -1
NOTE: When using these units it is important to distinguish between cal and Cal. The first is the normal calorie (sometimes called small calorie or gram-calorie), corresponding to the amount of heat required to raise the temperature of 1g of water, while Cal (with a capital letter) is a unit equivalent to 1,000 cal, or, what is the same, 1 kcal. This last unit of heat is used daily in health sciences, especially in the area of nutrition. In this context, it is the unit par excellence used to represent the amount of energy present in food (when we talk about calories in the context of food, we almost always mean Cal and not lime).
Examples of Specific Heat Calculation Problems
Below are two solved problems that exemplify both the process of calculating the specific heat for a pure substance and for a mixture of pure substances in which we know the specific heats.
Problem 1: Calculation of specific heat of a pure substance
Statement: You want to determine the composition of a sample of an unknown silver metal. It is suspected that it may be silver, aluminum or platinum. To determine what it is, the amount of heat required to heat a 10.0-g sample of the metal from a temperature of 25.0°C to the normal boiling point of water, that is, 100.0°C, is measured. obtaining a value of 41.92 cal. Knowing that the specific heats of silver, aluminum and platinum are 0.234 kJ.kg -1 .K -1 , 0.897 kJ.kg -1 .K -1 and 0.129 kJ.kg -1 .K -1 , respectively, Determine what metal the sample is made of.
What the problem asks is to identify the material from which the object is made. Since specific heat is an intensive property, it is characteristic of each material, so to identify it, it is enough to determine its specific heat and then compare it with the known values of the suspected metals.
The determination of the specific heat in this case is carried out by means of three simple steps:
Step #1: Extract all the data from the statement and carry out the relevant unit transformations
As in any problem, the first thing we need is to organize the data to have it at hand when needed. In addition, carrying out the unit transformations from the beginning will prevent us from forgetting it later and will also make the calculations easier in the following steps.
In this case, the statement gives the mass of the sample, the initial and final temperatures after a heating process, and the amount of heat required to heat the sample. It also gives the specific heats of the three candidate metals. In terms of units, we can note that the specific heats are in kJ.kg -1 .K .1 , but the mass, temperatures, and heat are in g, °C, and cal, respectively. We must then transform units so that everything is in the same system. It is easier to transform the mass, temperature and heat separately than to transform the compound units of the specific heat three times, so that will be the path we will follow:
Step #2: Use the equation to calculate the specific heat
Now that we have all the data we need, all we need to do is use the appropriate equation to calculate the specific heat. Given the data we have, we will use the second equation for Ce presented above.
Step #3: Compare the specific heat of the sample to the known specific heats to identify the material
When comparing the specific heat obtained for our sample with that of the three candidate metals, we observe that the one that most resembles it is silver. For this reason, if the only candidates are the metals silver, aluminum, and platinum, we conclude that the sample is composed of silver.
Problem 2: Calculation of specific heat of a mixture of pure substances
Statement: What will be the average specific heat of an alloy that contains 85% copper, 5% zinc, 5% tin, and 5% lead? The specific heats of each metal are, C e, Cu = 385 J.kg -1 .K -1 ; C e, Zn =381 J.kg -1 .K -1 ; C e, Sn = 230 J.kg -1 .K -1 ; C e, Pb = 130 J.kg -1 .K -1 .
This is a slightly different problem that requires a bit more creativity. When we have mixtures of different materials, the thermal properties and other properties will depend on the particular composition and, in general, will be different from the properties of the pure components.
Since specific heat is an intensive property, it is not an additive quantity, which means that we cannot add the specific heats to get a total specific heat for a mixture. However, what is additive is the total heat capacity, since this is an extensive property.
For this reason we can say that, in the case of the presented alloy, the total heat capacity of the alloy will be the sum of the heat capacities of the copper, zinc, tin and lead portions, that is:
However, in each case the heat capacity corresponds to the product between the mass and the specific heat, so this equation can be rewritten as:
Where C e al represents the average specific heat of the alloy (note that it is not correct to say total specific heat), that is, the unknown that we wish to find. As this property is intensive, its calculation will not depend on the amount of sample we have. In view of this, we can assume that we have 100 g of alloy, in which case the masses of each of the components will be equal to their respective percentages. By assuming this, we get all the data needed for the calculation of the average specific heat.
Now we substitute the known values and carry out the calculation. For simplicity, units will be ignored when substituting values. We can only do this because all specific heats are in the same system of units, as are all masses. It is not necessary to convert the masses to kilograms, since the grams in the numerator will eventually cancel out with those in the denominator.
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