Question

We mix 50.0 \(\mathrm{mL}\) of ethanol (density \(=0.789 \mathrm{g} / \mathrm{mL} )\) initially at \(7.0^{\circ} \mathrm{C}\) with 50.0 \(\mathrm{mL}\) of water \((\) density \(=1.0 \mathrm{g} / \mathrm{mL})\) initially at \(28.4^{\circ} \mathrm{C}\) in an insulated beaker. Assuming that no heat is lost, what is the final temperature of the mixture?

Short answer

The final temperature of the mixture can be calculated by equating the heat exchanged between water and ethanol considering their respective masses, specific heat capacities, and initial temperatures.

Step-by-step solution

Calculate Mass of Ethanol

Using the density formula \(\text{Density} = \frac{\text{mass}}{\text{volume}}\), calculate the mass of ethanol. With the given density of ethanol \(0.789 \mathrm{g} / \mathrm{mL}\) and its volume \(50.0 \mathrm{mL}\), the mass (m_ethanol) can be calculated as \(m_{\text{ethanol}} = \text{density of ethanol} \times \text{volume of ethanol} = 0.789 \mathrm{g/mL} \times 50.0 \mathrm{mL}\).

Calculate Mass of Water

Similarly, calculate the mass of water using its density. Since the density of water is \(1.0 \mathrm{g} / \mathrm{mL}\) and its volume is \(50.0 \mathrm{mL}\), the mass (m_water) is \(m_{\text{water}} = \text{density of water} \times \text{volume of water} = 1.0 \mathrm{g/mL} \times 50.0 \mathrm{mL}\).

Apply Conservation of Energy

Assuming no heat loss in the isolated system, apply the conservation of energy principle. The heat lost by the warmer substance (water) equals the heat gained by the cooler substance (ethanol): \(m_{\text{water}} \times C_{\text{water}} \times (T_{\text{final}} - T_{\text{water}}) = m_{\text{ethanol}} \times C_{\text{ethanol}} \times (T_{\text{ethanol}} - T_{\text{final}})\), where \(C_{\text{water}}\) and \(C_{\text{ethanol}}\) are the specific heat capacities of water and ethanol respectively, and \(T_{\text{final}}\), \(T_{\text{water}}\), and \(T_{\text{ethanol}}\) are the final, initial water, and initial ethanol temperatures respectively.

Solve for Final Temperature

After substituting known values (including \(C_{\text{water}} = 4.184 \mathrm{J/g^{\text{o}}C}\) and \(C_{\text{ethanol}} = 2.44 \mathrm{J/g^{\text{o}}C}\)), and solving for \(T_{\text{final}}\), we get a linear equation with one unknown, which can be solved algebraically to find the final temperature.

Key concepts

Conservation of Energy in Mixtures

Understanding the principle of conservation of energy in mixtures is crucial when predicting outcomes in a thermodynamic system. In the context of mixing substances, like ethanol and water, conservation of energy dictates that the total energy before mixing must equal the total energy after mixing, provided the system is isolated and no heat is lost to the surroundings.

This energy often manifests as thermal energy, and its transfer from one substance to another results in changes in temperature. In a mixture of two different temperatures, the hotter substance will lose heat energy, while the cooler substance gains it until thermal equilibrium is reached. It's important to remember that the heat lost by the hotter substance is equal to the heat gained by the cooler one. This understanding enables us to set up an equation that, when solved, tells us the final temperature of the mixture.

To improve comprehension, one could think of this as a ledger balance — if we 'lose' a certain amount of heat on one side, it must show up on the other to keep our 'energy account' balanced.

Specific Heat Capacity

The concept of specific heat capacity is at the heart of problems dealing with temperature changes within substances. It is defined as the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius.

Each substance has its own specific heat capacity, and it is a reflection of how much that material resists changing temperature. For example, water has a relatively high specific heat capacity, meaning it takes more energy to increase its temperature compared to a substance with lower specific heat capacity like ethanol. In our problem, knowing the specific heat capacities of both ethanol and water allows us to calculate how much heat energy is transferred to reach the final temperature of the mixture.

To make this more relatable, imagine specific heat capacity as a material's 'stubbornness', affecting how easy or difficult it is to change its temperature. When substances mix, they 'negotiate' based on their 'stubbornness' to settle at a final temperature that respects the conservation of energy.

Density Formula Application

The density formula application plays a foundational role in calculations involving mass and volume. Density is a property of matter that measures the mass of a substance per unit volume, which can be expressed as \( \text{Density} = \frac{\text{mass}}{\text{volume}} \).

In the problem at hand, the density of ethanol and water is used to calculate their respective masses from the given volumes. These masses are a necessary step, as energy changes in a mixture are tied to the mass of the substances involved. By applying the density formula, we convert volume measurements into mass, which then allows for the use of conservation of energy principles to find the final mixture temperature.

To visualize this, consider density as how tightly packed a substance's molecules are. In a practical sense, knowing how 'crowded' a substance is helps us to convert between how much space it takes up (volume) and how heavy it is (mass), enabling us to perform subsequent calculations. We can liken this technique to determining how many books you can fit in a box based on the space they occupy and their individual thickness.