Question

Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) boils at \(78.29^{\circ} \mathrm{C} .\) How much energy, in joules, is required to raise the temperature of \(1.00 \mathrm{kg}\) of ethanol from \(20.0^{\circ} \mathrm{C}\) to the boiling point and then to change the liquid to vapor at that temperature? (The specific heat capacity of liquid ethanol is \(2.44 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\) and its enthalpy of vaporization is \(855 \mathrm{J} / \mathrm{g} .\) )

Short answer

997,227 Joules.

Step-by-step solution

Convert Mass of Ethanol

To use specific heat and enthalpy equations, convert the mass of ethanol from kilograms to grams. Since 1 kg equals 1000 g, therefore:\[\text{Mass} = 1.00 \text{ kg} = 1000 \text{ g} \]

Calculate Energy to Heat Ethanol

Use the formula to calculate the energy required to raise the temperature of ethanol:\[ q = mc\Delta T \]where \(m = 1000 \text{ g}\), \(c = 2.44 \text{ J/g} \cdot \text{K}\), and the temperature change \(\Delta T = 78.29^{\circ} C - 20.0^{\circ} C = 58.29^{\circ} C\).\[ q = 1000 \text{ g} \times 2.44 \text{ J/g} \cdot \text{K} \times 58.29 \text{ K} = 142,227 \text{ J} \]

Calculate Energy for Vaporization

Use the enthalpy of vaporization to calculate the energy required to vaporize ethanol:\[ q_v = mH_v \]where \(m = 1000 \text{ g}\) and \(H_v = 855 \text{ J/g}\).\[ q_v = 1000 \text{ g} \times 855 \text{ J/g} = 855,000 \text{ J} \]

Total Energy Calculation

Sum the energy calculated for heating and vaporizing to find the total energy required:\[ q_{\text{total}} = q + q_v \]\[ q_{\text{total}} = 142,227 \text{ J} + 855,000 \text{ J} = 997,227 \text{ J} \]

Key concepts

Specific Heat Capacity

Specific heat capacity is an important concept when it comes to understanding how substances respond to changes in temperature. It is defined as the amount of heat energy required to raise the temperature of a given quantity of a substance by one degree Celsius (or one Kelvin).

In the context of this exercise, the specific heat capacity is used to determine the amount of energy needed to heat liquid ethanol. Here are a few things to understand about it:

• The specific heat capacity of ethanol, given as 2.44 J/g·K, implies that it takes 2.44 Joules to increase the temperature of one gram of ethanol by one Kelvin.
• This property varies from one substance to another, which is why it's crucial to use the correct value when performing energy calculations.
• Understanding this concept helps you predict how different substances react to heat and explains why some need more energy to change temperature.

Knowing the specific heat capacity allows us to use the formula \( q = mc\Delta T \), where \( q \) is the energy in Joules, \( m \) is the mass in grams, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change. This calculation gives insight into how much energy is needed for heating or cooling.

Energy Calculation

Energy calculations are a vital part of chemistry as they allow us to understand the amounts of energy involved in various physical and chemical processes. In this exercise, the energy calculation involves two parts: heating and vaporization.

To calculate the energy needed to heat the ethanol, we use the formula \( q = mc\Delta T \). This equation helps us determine the magnitude of energy required to increase the temperature of ethanol from 20.0°C to its boiling point 78.29°C:

• This involves calculating the change in temperature, \( \Delta T = 78.29 - 20.0 = 58.29 \) K.
• Applying the values, we get: \( q = 1000 \text{ g} \times 2.44 \text{ J/g} \cdot \text{K} \times 58.29 \text{ K} = 142,227 \text{ J} \).

Additionally, calculating the energy for vaporization is crucial:

• For this, we use the formula \( q_v = mH_v \), which calculates the energy needed to change the state from liquid to vapor.
• Given enthalpy of vaporization is 855 J/g, using \( m = 1000 \text{ g} \), the energy required is \( 855,000 \text{ J}. \)

The concept of energy calculation combines these processes to enhance our understanding of thermal dynamics in chemical reactions.

Temperature Change

Temperature change is a fundamental aspect when discussing energy transformations, especially in heat transfer. It's the driving factor in calculating how much energy is used or released in a substance.

In our exercise, understanding temperature change is crucial because:

• It informs the calculation of how much energy is needed to heat ethanol to its boiling point.
• The calculated temperature change, represented as \( \Delta T = 78.29^{\circ} C - 20.0^{\circ} C = 58.29 \) K, indicates how far the ethanol must be heated.
• By understanding \( \Delta T \), you can determine how quickly a substance can absorb or release heat, especially when transitioning between different states.

Often, temperature change is what leads to the phase change from liquid to vapor, requiring additional energy as calculated using the enthalpy of vaporization. This dual calculation showcases how temperature interacts with specific heat capacity and state changes to dictate the overall energy dynamics in a system.