Problem 116 - Chemistry An Atoms First Approach

Question

You have a 1.00 -L sample of hot water \(\left(90.0^{\circ} \mathrm{C}\right)\) sitting open in a \(25.0^{\circ} \mathrm{C}\) room. Eventually the water cools to \(25.0^{\circ} \mathrm{C}\) while the temperature of the room remains unchanged. Calculate \(\Delta S_{\text {surr }}\) for this process. Assume the density of water is \(1.00 \mathrm{g} / \mathrm{cm}^{3}\) over this temperature range, and the heat capacity of water is constant over this temperature range and equal to \(75.4 \mathrm{J} / \mathrm{K} \cdot\) mol.

Step-by-Step Solution

Answer

The entropy change for the surroundings during the process, as the hot water cools from 90.0°C to 25.0°C, is approximately 868.08 J/K.

Step 1: Calculate the mass of water

Given the volume of water as 1.00 L or 1000 cm³, we can use the density of water to determine the mass of water present in the sample. density = 1.00 g/cm³ volume = 1000 cm³ mass = density × volume = 1.00 g/cm³ × 1000 cm³ = 1000 g

Step 2: Convert mass of water to moles

We can convert the mass of water to moles using the molar mass of water, which is approximately 18.015 g/mol. mass_water = 1000 g molar_mass_water = 18.015 g/mol moles_water = mass_water / molar_mass_water = 1000 g / 18.015 g/mol ≈ 55.56 mol

Step 3: Calculate the heat transfer

To calculate the heat transfer from the hot water to its surroundings, we'll use the formula: q = n × C × ΔT where: - q is the heat transfer, - n is the number of moles, - C is the heat capacity of water (75.4 J/K·mol), and - ΔT is the change in temperature. ΔT = T_final - T_initial = 25.0°C - 90.0°C = -65.0 K Now, we can calculate the heat transfer: q = 55.56 mol × 75.4 J/K·mol × (-65.0 K) ≈ -258840 J

Step 4: Calculate the entropy change for the surroundings

To calculate the entropy change for the surroundings, we can use the formula: ΔS_surr = -q/T_surr where: - ΔS_surr is the entropy change of the surroundings and - T_surr is the temperature of the surroundings (25.0°C). First, convert the temperature of the surroundings to Kelvin: T_surr = 25.0°C + 273.15 = 298.15 K Now, we can calculate the entropy change for the surroundings: ΔS_surr = -(-258840 J) / 298.15 K ≈ 868.08 J/K So, the entropy change for the surroundings during this process is approximately 868.08 J/K.

Key Concepts

Heat Transfer in Thermodynamics

Heat transfer is a pivotal concept in thermodynamics that deals with the movement of energy from one place or material to another as a result of a temperature difference. In our everyday experience, we see heat transfer when a warm coffee cools down in a room: the heat energy is being transferred to the room. There are three primary methods of heat transfer: conduction, convection, and radiation.

Within a thermodynamic system like our exercise example, the system (hot water) and surroundings (the room) are exchanging energy. The water loses heat as it cools, which is absorbed by the ambient environment, thereby increasing the room's energy slightly. We express this heat transfer quantitatively using the formula:
\( q = n \times C \times Delta T \).

This equation leverages the number of moles of the substance (\(n\)), the heat capacity (\(C\)), and the temperature change (\(Delta T\)), highlighting how interrelated these concepts are in heat transfer calculations.

Molar Mass Calculation

In chemistry, the molar mass is the mass of one mole of a substance. It typically has units of grams per mole (\(g/mol\)) and is fundamental for converting between mass in grams and amount in moles. To find the molar mass, you would total the atomic masses of all atoms in the molecule, using the periodic table as a reference.

In our exercise, the molar mass calculation is critical for transforming the mass of water into the amount of substance in moles, which is an essential step for determining heat transfer. The molar mass of water, \(H_2O\), is about 18.015 \(g/mol\), derived from the atomic masses of hydrogen and oxygen; specifically, two atoms of hydrogen (approximately 1.008 \(g/mol\) each) and one atom of oxygen (approximately 15.999 \(g/mol\)).

Heat Capacity

The heat capacity of a substance is a measure of how much heat energy is needed to raise the temperature of a given amount of the substance by a certain temperature interval. Specifically, it is expressed in joules per kelvin per mole (\(J/K \text{\textbullet} mol\)). A material with a high heat capacity can absorb a lot of heat without a significant change in temperature, while a material with a low heat capacity heats up quickly.

In our textbook problem, we assume the heat capacity of water to be constant over the discussed temperature range. This simplification allows us to calculate the heat transferred as the water cools without considering any variations in heat capacity with temperature, which is typically a factor in more precise calculations.

Enthalpy and Entropy

Enthalpy and entropy are core concepts in thermodynamics, associated with heat content and disorder, respectively. Enthalpy (\(H\)) reflects the total heat content of a system and is concerned with heat transfer at constant pressure. Changes in enthalpy are used to determine whether a process is endothermic or exothermic.

Entropy (\(S\)), on the other hand, is a measure of randomness or disorder in a system. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. In our example, the entropy change calculation for the surroundings (\(Delta S_{\text{surr}}\)) is derived from the heat transferred to the room at a constant temperature, which indicates an increase in the room's entropy as the hot water cools and dispenses its heat.