# Understanding Coffee Cup Calorimetry

### Introduction to Coffee Cup Calorimetry

When I first heard about a calorimeter, I had no idea what it was or what relevance it would have to me and my major. Before I delve into the relevance of calorimetry though, let’s first talk about what it is. Calorimetry is the science of measuring the amount of heat transferred to or from a substance in a reaction by using a calorimeter to measure the heat exchanged with the surroundings. It is important to understand that in calorimetry problems, the substance reacting is the “system” and the water and calorimetry make up the “surroundings”. The “system” and “surroundings” exchange heat and this heat is what is measured.

A coffee cup calorimeter consists of a coffee cup, a thermometer, water, and a reactant placed inside the cup. Unlike a bomb calorimeter, a coffee cup calorimeter is a constant pressure calorimeter.

### Understanding Calorimetry

The basis of calorimetry is an assumption that the energy gained or lost by the surroundings is equal to the energy lost or gained by the system because heat is not lost the surrounding air. For example, if we were trying to determine the heat of fusion for ice melting inside a coffee cup calorimeter, we know that heat is only exchanged between ice and water. Therefore, we can assume that the energy gained by ice while melting is equal in magnitude but opposite in sign to the energy lost by the water inside the calorimeter.

$$\mathbf{Q}_{\mathrm{ice}}=-\mathbf{Q}_{\text { surroundings }}=-\mathbf{Q}_{\text { calorimeter }}$$

Another way to think about it is passing an apple to a friend. You give away the apple and have thus have -apple but your friend gains an apple and has +apple. You lost one apple and your friend gained one apple in the exchange (“equal in magnitude”) but you lost the apple and your friend gained the apple (“but opposite in sign”).

### Calorimetry Equation Basics

Now we can take a look at the main equation used in calorimetry problems and slowly go through each individual part of the equation.

$$\Delta \mathrm{Hr} \mathrm{x} \mathrm{n}=q_{\mathrm{rxn}}=-q_{\mathrm{calorimeter}}=-\mathrm{mC} \Delta \mathrm{T}$$

A coffee cup calorimeter is used to measure enthalpy changes in chemical processes, giving ΔH. Essentially, the heat measured in the device is equivalent to ΔH, the change in enthalpy. Since energy is neither created nor destroyed during a chemical reaction, the heat consumed or produced in the reaction, \( q_{\mathrm{reaction}}\) (the system), added to the heat lost or absorbed by the solution \(q_{\mathrm{solution}}\) (the surroundings) must sum to zero. If you’re feeling confused, think back to the ice and water example above and recall the apple analogy. It is the same as saying the heat gained by ice (the system) added to the heat lost by water (the surroundings) would equal zero because heat is only exchanged between the system and surroundings. In terms of \(q_{\text{reaction and solution}}\) we can visualize this concept as:

$$q_{\text { reaction }}+q_{\text { solution }}=0$$

In a reaction inside of a coffee cup calorimeter, the amount of heat released or absorbed by the calorimeter ( \(q_{\mathrm{solution}}\) ) is equal in magnitude and opposite in sign to the amount of heat consumed or produced by the reaction ( \( q_{\mathrm{reaction}}\) ).

$$q_{\text { reaction }}=-q_{\text { solution }}$$

Recall that the heat released or absorbed at constant pressure in a coffee cup calorimeter is equal to Δ*H*, the change in enthalpy. Thus the relationship between heat (\(c_{\text { alorimeter }}\)) and \(\Delta H_{\mathrm{rxn}}\) is

$$\Delta \mathrm{Hrxn}=q_{\mathrm{rxn}}=-q_{\text { calorimeter }}=-\mathrm{mC} \Delta \mathrm{T}$$

Note that heat capacity, C, can never be negative for a mass or a substance.

### Solving Coffee Cup Calorimetry Problems

Important things you need to know when solving a calorimetry problem include:

Heat capacity of the surroundings (C)

Masses of the system and the surroundings (m)

Temperature before and after reaction (ΔT)

**Let’s look at the following problem: **

**Attempting to find the specific heat of fusion of ice, you first take 30.0 grams of an ice cube and place it into a coffee cup calorimeter filled with 120.0 grams of water at 36.3 °C. A few minutes later, the ice block has melted completely and the water temperature has dropped to 19.2 °C. What is your experimental value for the specific heat of fusion of ice? (Note that “fusion” is the same as “melting”).**

To solve this problem, we should first recall that by placing an ice cube into water and having the ice cube melt means that the quantity of energy gained by the ice cube is equivalent to the energy lost by water:

$$\mathbf{Q}_{\mathrm{ice}}=-\mathbf{Q}_{\mathrm{calorimeter}}$$

In other words, the ice cube is gaining energy (positive) while the water inside the calorimeter is losing energy (indicated by the negative sign).

When setting up problems, I like to lay out what is given and what I am solving for. We are given the following:

**Mass of Ice**: 30.0 grams**Mass of Water inside Calorimeter:**120.0 grams**Initial Temperature:**36.3 °C**Final Temperature:**19.2 °C

We are solving for the following:

**Specific heat of fusion of ice**= ?

To start solving, we will need the main Calorimetry equation:

$$\Delta \mathrm{Hrxn}=\mathbf{q} \mathrm{rxn}=-\text { qcalorimeter }=-\mathrm{mC} \Delta T$$

Since we are solving for the heat of fusion for ice and we are given the mass of water, the change in temperature of the water before and after the ice cube melts, and we know the specific heat of water (which will always be 4.18 J/g/°C), we can solve for the heat **lost** by water, or \(\mathrm{Q}_{\text { calorimeter }}\). Plugging in what we know from the problem, we will get the following:

\(\mathbf{Q}_{\text { calorimeter }}\) = mass of water • specific heat of water • (Final temperature of water – Initial temperature of water)

\(\mathbf{Q}_{\text { calorimeter }}=\mathbf{m} \cdot \mathbf{C} \cdot \Delta \mathbf{T}\)

\(\mathbf{Q}_{\text { calorimeter }}=(120.0 \mathrm{g}) \cdot\left(4.18 \mathrm{J} / \mathrm{g} /^{\circ} \mathrm{C}\right) \cdot\left(19.2^{\circ} \mathrm{C}-36.3^{\circ} \mathrm{C}\right)\)

\(\mathbf{Q}_{\text { calorimeter }}=-8577.36 \mathrm{J}\)

The negative value of \(\mathbf{Q}_{\text { calorimeter }}\) supports what we already know—ice melting results in a gain of energy in the ice (system) and a loss of energy in the water (surroundings). We also assume that the energy lost by water will be gained in equal magnitude but in opposite sign to ice. Therefore, \(Q_{\text { ice }}=+8577.36 \mathrm{J}\) with the positive sign indicating a gain in energy** **for the ice.** **Referring to the calorimetry equation, we can calculate \(\Delta \mathrm{H}_{\text { fusion-ice }}\) using this set up:

Heat gained by ice = Mass of ice • Specific heat of fusion of ice

\(\mathbf{Q}_{\mathrm{ice}}=\mathrm{m}_{\mathrm{ice}} \bullet \Delta \mathrm{H}_{\mathrm{fusion}-\mathrm{ice}}\)

One question you might ask is if we are using the calorimetry equation to solve for \(\Delta \mathrm{H}_{\mathrm{fusion} \text { ice }}\), then why isn’t the change in temperature included in the equation above? One important thing to remember is our phase change diagrams.

As ice gains energy in the form of heat, its temperature will increase until it reaches its melting point. Once ice reaches its melting point, it will begin to melt. As ice begins to melt to liquid water, the temperature remains constant even though heat is being added to the system (the ice). The temperature will remain constant until all the ice has melted into liquid water. On the graph shown above, the position of ice melting is at a flat line denoted by a green dot. Through this graph we visually see that although energy is increasing in the system, temperature stays constant as ice melts.

Now that we have clarified why change in temperature is not included in the equation, we can continue solving the problem to find the heat of fusion of our ice cube:

\(\mathbf{Q}_{\mathrm{ice}}=\mathrm{m}_{\mathrm{ice}} \bullet \Delta \mathrm{H}_{\mathrm{fusion}-\mathrm{ice}}\)

\(+8577.36 \mathrm{J}=(30.0 \mathrm{g}) \cdot \Delta \mathrm{H}_{\text { fusion-ice }}\)

\(\Delta \mathrm{H}_{\text { fusion-ice }}=(+8577.36 \mathrm{J}) /(30.0 \mathrm{g})\)

\(\Delta \mathrm{H}_{\text { fusion-ice }}=285.91 \mathrm{J} / \mathrm{g}\)

\(\Delta \mathrm{H}_{\text { fusion-ice }}=2.86 \mathrm{x} 10^{2} \mathrm{J} / \mathrm{g}\) (rounded to two significant figures)