## Section24.2Energy Conservation in Calorimetry

What will happen if you drop a hot copper block in water at room temperature? Energy from hot copper block will flow into water and raise the temperature of water. First law of thermodynamics helps us address this quantitatively and figure out the final state in this situation. Let us address this type of objects coming in thermal contact in more general terms first.

When two systems at different temperatures come in a thermal contact, energy is exchanged between them through heat and work. The energy conservation says that if the two systems together are isolated from everything else, then their combined energy will remain the same, and any gain in the energy of one will be accompanied by equal loss of energy by the other. Labeling the two systems as A and B we say that

$$\Delta E_A = -\Delta E_B.\label{eq-calor-en-bal-1}\tag{24.2.1}$$

We classify the transfer of energy from one to the other as heat $Q$ and work $W\text{.}$ Let $Q_A$ be the amount of heat into A and $W_A$ be the amount of work on A. Then, the change of energy of A is

\begin{equation*} \Delta E_A = Q_A + W_A. \end{equation*}

Similarly, suppose $Q_B$ be the amount of heat into B and $W_B$ be the amount of work on B. Then,

\begin{equation*} \Delta E_B = Q_B + W_B. \end{equation*}

Writing Eq. (24.2.1) in terms of heat and work we obtain

$$Q_A + W_A = -\left(Q_B + W_B\right). \label{eq-calor-en-bal-2}\tag{24.2.2}$$

Now, if $W_A$ and $W_B$ are negligible, then this balancing of energy leads to the balancing of heat into A and heat into B.

$$Q_A \approx - Q_B. \label{eq-calor-en-bal-3}\tag{24.2.3}$$

On the right side we have the negative of the heat into B, which is equal to the heat out of B. Therefore, we can say that if neither of the two interacting systems do work, then the heat into one of the systems is equal to the heat out of the other system. We write this result in more colloquial language as

$$|\text{Heat Lost}| \approx |\text{Heat Gained}|\ \ \ \ (\text{since}, W_{\text{on either system}} \approx 0.) \label{eq-calometry-1}\tag{24.2.4}$$

While working out the heat gained or the heat lost, you must pay particular attention to any phase change(s) and not only to the changes in the temperature. At each phase change you will need to account for the heat of transformation involved in the change of phase. I will illustrate these cautionary remarks in the examples below.

A more serious limitation in the use of Eq. (24.2.4) is that this equation is valid only when no work is done by either system. Therefore, we normally use Eq. (24.2.4) for experiments with either liquids or solids since they are somewhat incompressible. This equation can also be applied to a system of gas in a rigid container to prevent the volume of the gas from changing. If volume of the system changes much, then Eq. (24.2.4) cannot be used since energy conservation will not work if you take into account only the heat part, you must also include the amount in work.

A device using the principles of heat exchange in the absence of any work can be used to measure the heat released in a particular process. Such a device is called a calorimeter, and the technique of measuring heat this way is called calorimetry. We will illustrate the idea with examples.

### Subsection24.2.1Exercises

Follow the link: Calirometry in Thermal Properties Chapter 22.3.