Problem 20 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The molar heat capacity of water at constant pressure is \(75.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) at room temperature. Calculate the mass of methanol that must be burned to heat \(1 \mathrm{dm}^{3}\) of water from \(20^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). (Assume the density of water is \(1.0 \mathrm{gcm}^{-3}\) ) (Section \(13.3)\)

Step-by-Step Solution

Answer

Burn ~5.53 g of methanol to heat the water.

Step 1: Calculate the Mass of Water

Given that the density of water is \(1.0\,\text{g/cm}^3\) and the volume is \(1\,\text{dm}^3\), we can convert to grams since \(1\,\text{dm}^3 = 1000\,\text{cm}^3\). Thus, the mass of the water is \(1000\,\text{g}\).

Step 2: Calculate the Moles of Water

Convert the mass of water to moles using the molar mass of water \(\approx 18.015\,\text{g/mol}\). Calculate: \( \text{moles of water} = \frac{1000\,\text{g}}{18.015\,\text{g/mol}} \approx 55.5\,\text{mol} \).

Step 3: Calculate the Heat Required

Use the formula for heat \( q = nC_p\Delta T \). Given \( C_p = 75.3\,\text{J/K/mol} \) and \( \Delta T = 50^{\circ} \text{C} - 20^{\circ} \text{C} = 30\,\text{K} \), calculate: \( q = 55.5\,\text{mol} \times 75.3\,\text{J/K/mol} \times 30\,\text{K} = 125252.25\,\text{J} \).

Step 4: Calculate Energy Released by Combustion of Methanol

The combustion of methanol releases \(-726\,\text{kJ/mol}\). Convert \(125252.25\,\text{J}\) to \(\text{kJ}\): \(125.252\,\text{kJ}\). Determine the moles of methanol needed: \( \text{moles of methanol} = \frac{125.252\,\text{kJ}}{726\,\text{kJ/mol}} \approx 0.1725\,\text{mol} \).

Step 5: Calculate Mass of Methanol

Use the molar mass of methanol \(\approx 32.04\,\text{g/mol}\) to find the mass: \( \text{mass of methanol} = 0.1725\,\text{mol} \times 32.04\,\text{g/mol} \approx 5.53\,\text{g} \).

Key Concepts

Molar Heat Capacity

Molar heat capacity is a fascinating concept in thermochemistry. It's the amount of heat needed to raise the temperature of one mole of a substance by one Kelvin (or one degree Celsius).
For water, at room temperature, this value is typically about 75.3 J/K/mol.
This high molar heat capacity means water can absorb a lot of heat without a significant temperature change, which is why it's such an effective coolant.
In practical applications, knowing the molar heat capacity is crucial for calculating how much energy is required to heat a given amount of a substance.

• To find the total energy needed, multiply the molar heat capacity (Cp) by the number of moles and the temperature change (ΔT).
• This is expressed in the equation: \( q = nC_p \Delta T \) where \( q \) is the heat energy.

Remember, this calculation assumes pressure is constant, which is a common condition in many real-world scenarios.

Combustion of Methanol

The combustion of methanol (CH3OH)is a chemical reaction where methanol reacts with oxygen to produce carbon dioxide, water, and energy.
The process releases significant energy, specifically 726 kJ per mole of methanol combusted. This energy release makes methanol a useful fuel source. It's critical to balance energy release with the energy needed for specific applications.
In experiments or calculations, you'll often compare the energy from methanol combustion to the energy required in a process like heating water.

• The formula for combustion of methanol is: \[2 ext{CH}_3 ext{OH} + 3 ext{O}_2 \rightarrow 2 ext{CO}_2 + 4 ext{H}_2 ext{O} \]
• Understanding this reaction is essential for tasks like calculating the mass of methanol needed to produce a certain amount of heat energy.

So, whether using it for fuel or for heat transfer calculations, knowing about methanol combustion is vital.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics. It's the movement of thermal energy from one object or substance to another. This process occurs in three main modes: conduction, convection, and radiation.
In the context of heating water, primarily conduction is involved when energy is transferred from a heat source to water molecules.
A key part of solving real-world heat transfer problems is understanding how much energy is transferred and how it affects temperature changes. Common equations like \(q = nC_p \Delta T\) are used in calculations to determine the total heat transfer involved.

• In simple terms, if you want to increase the temperature of a substance, you need to add energy equal to the heat capacity times the temperature change.
• This principle is used in designing everything from domestic water heaters to large-scale industrial processes.

By mastering heat transfer, you'll better understand how temperature changes in substances and how energy conservation applies to thermochemical reactions.

Density of Water

The density of water is a fundamental physical property, often used as a benchmark in scientific calculations. It's around 1.0 \( ext{g/cm}^3 \) at room temperature and essential for converting between volume and mass.
Knowing this density allows easy calculation of the mass of water from its volume and vice versa.
In practical applications, such as heating water, you often need the exact water mass to accurately determine the required energy.

• For example, if you have \( 1 ext{ dm}^3 \) (which is \( 1000 ext{ cm}^3 \),) of water, the mass is \( 1000 ext{ g} \).
• This conversion is crucial when applying formulas that involve mass, such as calculating the moles of water for heat capacity calculations.

Understanding the density of water helps you relate its physical quantity with the energy changes involved in heating or cooling processes efficiently.