Heat capacity


Since temperature is a measure of the average amount of thermal energy in an object, an object that absorbs heat increases its temperature. But by how much is not necessarily obvious. In the early days of the study of heat, it was thought that heat was a kind of a fluid and that the object's temperature measured how much of that fluid was in the object. (See the Wikipedia article on phlogiston for details.) This sounds similar to our understanding of higher temperature as being associated with more energy, but it's not the same. We will learn when we study microscopic models of matter that temperature is related to the average energy in a degree of freedom  -- a place that energy can reside: the kinetic energy of the motion of a molecule in the x direction or the potential spring energy of a bond between two molecules in a solid.

The important implication of this is that because objects are made of different materials, the amount of heat required will vary depending on how hard it is to get the molecules to increase their motion and hence their temperature. Or in other words:

Every material translates energy into temperature in its own way.

We'll see in our discussion of the microscopic structure of matter exactly how this works. But for now, let's only consider the macroscopic phenomenology. For a particular object, we can measure the amount of heat necessary to raise that object’s temperature by one degree (K).  This is the object's heat capacity, C.  If there is no significant heat loss in the time we are considering (see Heat Transfer) then the heat needed to make the temperature increase by an amount $\Delta T$ is

$$Q = C \Delta T$$                          

The heat capacity can give us good insight into how heat tends to be distributed.  If we consider two objects labeled 1 and 2 with different initial temperatures and different specific heats, and we allow them to exchange energy until they reach the same temperature, that temperature will be given by

$$Q_1 = -Q_2$$

$$Q_1 = C_1 \Delta T_1\quad \quad Q_2 = C_2 \Delta T_2$$     


$$ C_1 ( T^f_1 - T^i_1) = -C_2 ( T^f_2 - T^i_2)$$

Setting the initial temperatures of objects 1 and 2 to $T_1$ and $T_2$ respectively, and setting both final temperatures equal to the same value, $T$, we can solve for $T$ to get

$$T = \bigg(\frac{C_1}{C_1+C_2}\bigg)T_1  +\bigg( \frac{C_2}{C_1+C_2}\bigg)T_2$$

This looks sensible. The final temperature is the average of the two initial temperatures weighted by the fraction of the total specific heat in each object.

Empirically, this is a pretty good model over a fairly wide range of everyday temperatures: the change in temperature of an object due to heat transfer is directly proportional to the amount of heat in or out of the object. But for large temperature changes, the dependence can become more complex. (Engineers need this for steam engines.) The way this is handled is by making the heat capacity temperature dependent. For biological systems, the typical ranges of temperature are fairly small so we don't have to worry about this.

If the object consists of a single uniform material, each gram of the object responds to heat in the same way. We can therefore introduce a thermal response quantity for a material that is density-like: the amount of heat required to raise 1 kg of that material by one degree. This is called the specific heat, c. The units of c in SI units will be J/kg-K. The heat capacity of a uniform object is therefore its mass times the specific heat of the material it's made of:

$$C = mc$$                         

We can use heat capacity to figure out how much energy is needed to increase an object’s temperature by any amount. 

An interesting example would be the amount of energy that your body uses to raise your temperature by 1 K, say when you have a slight fever. Although you are not uniform, most of you is water, so we can assume that the heat capacity of a person is the same as water, $4.2 \times 10^3$ J/kg K. If $Q$ is the amount of heat and $ΔT$ is the temperature change, for a 150 lb person (70 kg) the heat needed is

  $$Q = mc \Delta T  = \bigg(4.2 \times 10^3 \frac{\mathrm{J}}{\mathrm{kg\;K}}\bigg) \times (70\;\mathrm{kg}) \times 1 \mathrm{K} \approx 3 \times 10^5 \mathrm{J}$$

or about 300 kJ. One way is to think about how this energy is to compare it to that coming from a 100 W light bulb, which emits 100 J of energy per second.  (This refers to old-style, incandescent bulbs, which you no longer see very often.) This amount of energy is equivalent to the energy emitted over about an hour. 

Workout: Heat capacity



Julia Gouvea and Joe Redish 4/27/15


Article 435
Last Modified: September 22, 2021