# 5.2: Heat Capacity


Learning Objectives

• Define heat capacity and specific heat capacity and differentiate between the two terms
• Deduce which substance will have greatest temperature changed based on specific heat capacities
• Calculate unknown variables based on known variables using the specific heat equation

## Heat Capacitance

The heat capacity of a substance describes how its temperature changes as it absorbs or releases heat, it is the capacity of a substance to contain heat. This equation relates the heat absorbed (or lost) to the temperature change

$\Delta q = q_{\text{transferred}}=q =C\Delta T$

Please note this heat q is the amount of heat transferred to or from an object as its temperature is changed, and we call it q, not (\Delta q\). It is not the total heat energy of the object.

Rearranging the above Equation gives the definition of the heat capacitance of an object $C=\frac{q(J)}{\Delta T(K)}$

This is an important equation because it relates the heat lost or gained by an object to its temperature change. A substance with a small heat capacitance cannot hold a lot of heat energy and so warms up quickly. We should note that the rate of heat transfer (how fast something heats up or cools down), is a function of the temperature difference. So as heat transfers from the hot object (which cools) to the cold object (which warms) the temperature difference reduces, and the rate of heat transfer slows down and finally stops when they reach the same temperature. The heat capacitance of a substance depends both on the material it is made of, and the mass of the substance.

Note: You can determine the above equation from the units of Capacity (energy/temperature). That is if a constant has units, the variables must fit together in an equation that results in the same units. So C equals something with energy in the numerator and temperature in the denominator. Now, you need to use some common sense here, as we are adding heat, not work, and adding heat changes the temperature, it does not make the temperature. So the right side is a ΔT, and not a T.

Exercise $$\PageIndex{1}$$

The Heat Capacitance of a 10.0 g silver ring is 2.36J/oC.

1. Convert this value to units of J/K.
2. Convert this value to units of cal/oC.

2.36J/K, these are the same value because the denominator is $$Delta T$$ not T.

0.564cal/oC.

## Specific Heat Capacitance

The specific heat capacitance is the heat capacitance per gram of a substance. This value depends on the nature of the chemical bonds in the substance, and its phase.

$q = mc\Delta T$

or

$c=\frac{q(J)}{m(g)\Delta T(K)}$

Note: Capital "$$C$$" is the Heat Capacity of an object, lower case "$$c$$" is the specific heat capacity of a substance. The heat capacity of an object made of a pure substance is:

$C=mc$

If the material of an object is made of uniform in composition you can use the specific heat capacity for that material to calculate the heat capacitance of the object. So doubling the mass of an object doubles its heat capacity, but does not change its specific heat capacitance.

Exercise $$\PageIndex{2}$$

In exercise $$\PageIndex{1}$$ we saw that a 10.0 g silver ring has a heat capacitance of 2.36J/oC, what is the specific heat capacity of silver?

Assuming the ring is pure silver, the specific heat capacity of silver is 0.236J/goC.

Tip: Using Units of a Constant to the Identify equation

It should be noted that just as for heat capacity, the units of specific heat capacity must align with the units of the equation, and so you can calculate the equation from the units, as long as you realize J is a unit of energy (we are talking heat, not work), g is a unit of mass, and °C is a unit of temperature, although here, it stand for temperature change (ΔT). Video $$\PageIndex{1}$$ shows how we can use the units of a constant to determine the equation.

Is specific heat capacitance an extensive or intensive property?

The specific heat capacity is intensive, and does not depend on the quantity, but the heat capacity is extensive, so two grams of liquid water have twice the heat capacitance of 1 gram, but the specific heat capacity, the heat capacity per gram, is the same, 4.184 (J/g.K). So a table of specific heat capacitance based on the type of material can be used to allow us to calculate the heat capacitance of an object. Note the heat capacity depends on the phase of the substance.

Table $$\PageIndex{1}$$: Specific Heat Capacities for common substances.
Substance Symbol (state) Specific Heat (J/g °C)   Substance Symbol (state) Specific Heat (J/g °C)
helium He(g) 5.193   aluminum Al(s) 0.897
water H2O(l) 4.184   carbon dioxide CO2(g) 0.853
ethanol C2H6O(l) 2.376   Silicon Si(s) 0.712
ice H2O(s) 2.093 (at −10 °C)   argon Ar(g) 0.552
water vapor H2O(g) 1.864   iron Fe(s) 0.449
nitrogen N2(g) 1.040   copper Cu(s) 0.385
air mixture 1.007   gold Au(s) 0.129
oxygen O2(g) 0.918   lead Pb(s) 0.128

Note: Metals have low heat capacities and thus undergo rapid temperature rises when heat is applied.

Exercise $$\PageIndex{3}$$

If you add the same amount of heat to an equal mass of liquid water, solid gold, and solid iron, which would end up having the highest temperature?

Solid Gold. They all have the same mass and are exposed to the same amount of heat. So, the one with the lowest specific heat would have the highest temperature. It has the lowest resistance to temperature change when exposed to heat. If you ever reached into an oven to grab your food with a gold bracelet on, you may have experience the low specific heat capacity of gold. Metals have low heat capacities and thus undergo rapid temperature rises when heat is applied.

Example $$\PageIndex{1}$$

What is the final temperature if 100.0 J is added to 10.0 g of aluminum at 25oC? $$c){Al} = 0.902\,J/(g.^oC)$$. For extra help watch video $$\PageIndex{2}$$, which solves this problem.

Solution

$T_f=T_i+\frac{q}{mc}=25^oC+\frac{100.0\,J}{10.0\,g(\frac{0.902\,J}{g \cdot ^0C})}=36.1^oC \nonumber$

### Specific heat capacity can be used to identify an unknown substance

The specific heat capacity is a physical property of the material a substance is composed of and can be used to help identify the substance the way density can help identify an incompressible substance like a solid or liquid. It should be noted that two substances can have the same specific heat capacity just as two substances can have the same density, but for example, if the heat capacity of a clear liquid is not 1 cal/goC, the substance can not be pure water.

Example $$\PageIndex{2}$$

You have an unknown metal that is either Al, Cu, Ag or Fe and want to identify it. Upon adding 51.26J to 10.0G of the metal its temperature is raised by 22oC

Solution

Calculate the specific heat capacity and relate to those on table Table $$\PageIndex{1}$$. The metal is silver and this problem is solved in Video $$\PageIndex{3}$$

Exercise $$\PageIndex{4}$$

Can you use the specific heat capacity to tell the difference between lead and gold?

No, gold had a specific heat capacity of 0.129J/goC and lead of 0.128J/goC. The thousandths position is uncertain and so to three significant digits, you can not differentiate between these two samples (you report all certain and the first uncertain, so if you have a measurement of 0.128, you are not really sure of the 0.008 value.