Question

Using information in Appendices \(\mathrm{B}\) and \(\mathrm{C}\), calculate the minimum number of grams of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\), that must be combusted to provide the energy necessary to convert \(5.50 \mathrm{~kg}\) of ice at \(-20^{\circ} \mathrm{C}\) to liquid water at \(75^{\circ} \mathrm{C}\).

Short answer

The minimum amount of propane needed to convert \(5.50\,\mathrm{kg}\) of ice at \(-20^{\circ}\mathrm{C}\) to liquid water at \(75^{\circ}\mathrm{C}\) is \(75.14\,\mathrm{g}\).

Step-by-step solution

Energy required to raise the temperature of ice

Firstly, we calculate the energy required to raise the temperature of 5.50 kg ice at -20°C to 0°C. The formula for energy required (Q) is: \(Q = mcΔT\) Where, m = mass of the ice (5.50 kg), c = specific heat capacity of ice = 2.093 \(\frac{kJ}{kg \cdot °C}\), ΔT = change in temperature = 20°C \(Q_1 = (5.50\,\mathrm{kg})(2.093\,\frac{kJ}{kg \cdot °C})(20\,°C) = 229.2\,\mathrm{kJ}\) So, 229.2 kJ energy is required to raise the temperature of ice from -20°C to 0°C. #Step 2: Energy to convert ice to water at 0°C#

Energy required to melt the ice

Next, we need to calculate the energy required to convert the ice at 0°C to water at 0°C. The formula for energy required (Q) for a phase change is: \(Q = mL\) Where, L = heat of fusion for ice = 334 \(\frac{kJ}{kg}\) \(Q_2 = (5.50\,\mathrm{kg})(334\,\frac{kJ}{kg}) = 1837\,\mathrm{kJ}\) So, 1837 kJ energy is required to melt the ice at 0°C. #Step 3: Energy to raise the temperature of water to 75°C#

Energy required to raise the temperature of water

We now need to calculate the energy required to raise the temperature of water from 0°C to 75°C. The specific heat capacity of water = 4.186 \(\frac{kJ}{kg \cdot °C}\), ΔT = change in temperature = 75°C \(Q_3 = (5.50\,\mathrm{kg})(4.186\,\frac{kJ}{kg \cdot °C})(75\,°C) = 1716.1\,\mathrm{kJ}\) So, 1716.1 kJ energy is required to raise the temperature of water to 75°C. #Step 4: Calculate the total energy required#

Total energy required for the entire process

Add the energy required for each step: \(Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 229.2\,\mathrm{kJ} + 1837\,\mathrm{kJ} + 1716.1\,\mathrm{kJ} = 3782.3\,\mathrm{kJ}\) So, a total of 3782.3 kJ energy is required for the entire process. #Step 5: Calculate the energy released by combustion of 1g of propane#

Energy released by combustion of 1g of propane

We will use the heat released from the combustion of propane, which is 2220 kJ/mol. To find the heat released per gram, we need to divide it by the molar mass of propane. Molar mass of propane, \(\mathrm{C}_{3}\mathrm{H}_{8}\) = 44.1 g/mol Energy released per gram of propane: \(\frac{2220\,\mathrm{kJ/mol}}{44.1\,\mathrm{g/mol}} = 50.34\,\frac{kJ}{g}\) #Step 6: Determine the minimum amount of propane required#

Minimum amount of propane required

Now that we know the energy required and the energy released per gram of propane, we can calculate the minimum amount of propane needed. Min amount of propane = \(\frac{\text{Total energy required}}{\text{Energy released per gram of propane}}\) Min amount of propane = \(\frac{3782.3\,\mathrm{kJ}}{50.34\,\frac{kJ}{g}} = 75.14\,\mathrm{g}\) So, the minimum amount of propane needed is 75.14 grams.

Key concepts

Specific Heat Capacity

The concept of specific heat capacity is critical when it comes to understanding heat transfer in materials. It represents the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). This property varies among different substances and is vital in various scientific calculations related to temperature changes.

The expression for calculating the energy, Q, associated with a change in temperature is given by the formula:
\[Q = mc\Delta T\]
where:

• \(m\) is the mass of the substance,
• \(c\) is the specific heat capacity, and
• \(\Delta T\) is the change in temperature.

In the exercise, we first encountered specific heat capacity when calculating the amount of energy required to raise the temperature of ice. It's clear from the solution that a substantial amount of energy, 229.2 kJ, is required just to bring ice from -20°C to 0°C. This demonstrates how specific heat capacity is an integral part of thermal energy calculations, aiding in understanding how much energy is involved in heating or cooling a substance.

Heat of Fusion

When matter changes from one phase to another, like from solid to liquid, the energy involved in this process is quantified by the heat of fusion. Specifically, the heat of fusion refers to the amount of energy necessary for a substance to change from the solid phase to the liquid phase at its melting point without changing its temperature.

To calculate this energy, we use the formula:\[Q = mL\]
where:

• \(L\) represents the heat of fusion of the substance, and
• \(m\) is the mass undergoing the phase change.

For our ice-to-water transition at 0°C, we needed to understand the heat of fusion to calculate the energy required to melt the ice. As evidenced in the solution, 1837 kJ of energy is solely for the phase change, with no temperature change. Understanding the heat of fusion is vital for industries like food processing and metallurgy, where controlled phase changes are frequent.

Enthalpy of Combustion

Enthalpy of combustion is a term used to describe the heat released when a substance completely reacts with oxygen under standard conditions. It's typically expressed in kJ/mol and is a crucial consideration when assessing the energy content of fuels.

The enthalpy of combustion can be connected to real-life applications such as calculating the amount of a fuel required to produce a certain amount of energy. This is exactly what we did in the problem when we calculated how much propane was needed to provide enough energy to convert ice to water at a higher temperature.

To find the energy per gram of propane released during combustion, we divided the enthalpy of combustion by the molar mass of propane:\[\frac{2220\,\mathrm{kJ/mol}}{44.1\,\mathrm{g/mol}} = 50.34\,\frac{kJ}{g}\]
This resulted in requiring 75.14 grams of propane to supply the 3782.3 kJ of energy for the entire heating process. Understanding the enthalpy of combustion is fundamental to energy management and conservation strategies in various industries, including automotive, heating, and power generation.