Question

Methanol can be used in automobiles as antifreeze. Calculate the heat required to vaporize \(5.00 \mathrm{~L}\) of the liquid at \(25.0^{\circ} \mathrm{C}\). (For methanol: density \(=0.792 \mathrm{~g} / \mathrm{mL} ;\) specific heat \(=0.610 \mathrm{cal} /\left(\mathrm{g} \times{ }^{\circ} \mathrm{C}\right) ; \mathrm{Bp}=65.0{ }^{\circ} \mathrm{C} ; \mathrm{H}_{\text {vapor }}=\) \(263 \mathrm{cal} / \mathrm{g} .)\),

Short answer

1138224 cal is required.

Step-by-step solution

Convert Volume to Mass

First, we need to convert the volume of methanol from liters to milliliters. Since 1 liter equals 1000 milliliters, the volume is \(5.00 \, \text{L} = 5000 \, \text{mL}\). Next, we use the density of methanol to find the mass. Using the formula \( \text{mass} = \text{density} \times \text{volume} \), we get: \( \text{mass} = 0.792 \, \text{g/mL} \times 5000 \, \text{mL} = 3960 \, \text{g}\).

Heat Required to Raise Temperature

Calculate the heat required to raise the temperature of methanol from \(25.0^{\circ} \text{C}\) to its boiling point at \(65.0^{\circ} \text{C}\) using the specific heat capacity. The formula is: \( q_1 = \text{mass} \times \text{specific heat} \times \Delta T\).\( \Delta T = 65.0 - 25.0 = 40.0^{\circ} \text{C} \). Thus, \( q_1 = 3960 \, \text{g} \times 0.610 \, \text{cal/g}^{\circ} \text{C} \times 40.0^{\circ} \text{C} = 96744 \, \text{cal}\).

Heat Required to Vaporize

Now calculate the heat needed to vaporize the methanol using its heat of vaporization. The formula is \( q_2 = \text{mass} \times \text{H}_{\text{vapor}} \).\( q_2 = 3960 \, \text{g} \times 263 \, \text{cal/g} = 1041480 \, \text{cal}\).

Total Heat Required

Add the heat calculated in Step 2 and Step 3 to find the total heat required.\( q_{\text{total}} = q_1 + q_2 = 96744 \, \text{cal} + 1041480 \, \text{cal} = 1138224 \, \text{cal}\).

Key concepts

Vaporization

Vaporization is the process where a substance changes from a liquid to a gas. For methanol, vaporization occurs when it is heated to its boiling point. The key factor in calculating the heat required for vaporization is the heat of vaporization. This value represents the amount of energy needed to vaporize one gram of a substance, in this case, methanol, which is given as 263 cal/g. To find the total energy needed for vaporizing a specific mass of methanol, you simply multiply the mass by the heat of vaporization using the formula:

• \( q_2 = \text{mass} \times \text{H}_{\text{vapor}} \)

This step is crucial for determining the heat required during phase changes, where temperature remains constant until the substance fully transitions.

Specific Heat Capacity

Specific heat capacity is a property that defines how much heat is required to change the temperature of a one-gram sample of a substance by one degree Celsius (°C). For methanol, the specific heat capacity is 0.610 cal/(g°C). This value allows us to calculate the heat needed to raise the temperature of a given mass of methanol before it reaches its boiling point. The formula used is:

• \( q_1 = \text{mass} \times \text{specific heat} \times \Delta T \)

where \( \Delta T \) is the temperature change in °C. Knowing the specific heat helps in determining how much energy is required to heat the substance to the point where it can begin to vaporize.

Temperature Change

Temperature change is often a critical component when calculating heat transfer in substances. It is simply the difference between the final and initial temperatures of the substance. In this scenario, methanol's temperature must increase from 25.0°C to 65.0°C to reach its boiling point, resulting in a temperature change (ΔT) of 40.0°C. The formula used to calculate the heat required for this change, combined with specific heat capacity, emphasizes how different substances absorb and retain heat differently.

• \( \Delta T = T_{\text{final}} - T_{\text{initial}} \)

This value helps dictate the amount of energy required to move methanol up to the point of vaporization.

Density Conversion

Density conversion is pivotal when shifting between units of volume to units of mass. Density refers to the mass per unit volume of a substance, expressed here as 0.792 g/mL for methanol. When tasked with calculating the mass from a given volume, the formula is:

• \( \text{mass} = \text{density} \times \text{volume} \)

In our case, converting 5.00 L of methanol to grams involves first converting liters to milliliters—because 1 L equals 1000 mL—and then applying the density to determine the mass. This conversion is foundational to quantifying how much of a substance you have before performing further heat calculations.