Question

calculate the amount of heat needed to boil 75.8 g of ethanol (ch₃ch₂oh), beginning from a temperature of −13.1 °c. round your answer to 3 significant digits. also, be sure your answer contains a unit symbol.

Explanation:

To solve this, we need the following data (assumed or standard values for ethanol):

• Melting point of ethanol: \(-114^\circ\text{C}\) (but we start at \(-13.1^\circ\text{C}\), so we only heat liquid ethanol to its boiling point, then vaporize it? Wait, no—wait, the problem says "boil" ethanol, so we need to:

1. Heat liquid ethanol from \(-13.1^\circ\text{C}\) to its boiling point (\(T_b = 78.4^\circ\text{C}\)) using \(q_1 = mc\Delta T\).
2. Vaporize the liquid ethanol at its boiling point using \(q_2 = m\Delta H_{\text{vap}}\).

Step 1: Identify constants (standard values for ethanol):

• Specific heat of liquid ethanol, \(c = 2.46\ \text{J/g}^\circ\text{C}\)
• Molar mass of ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)): \(M = 46.07\ \text{g/mol}\)
• Enthalpy of vaporization, \(\Delta H_{\text{vap}} = 38.56\ \text{kJ/mol}\) (or \(38560\ \text{J/mol}\))

Step 2: Calculate \(q_1\) (heating liquid ethanol to boiling point):

\(\Delta T = T_b - T_{\text{initial}} = 78.4^\circ\text{C} - (-13.1^\circ\text{C}) = 91.5^\circ\text{C}\)
\(q_1 = mc\Delta T = 75.8\ \text{g} \times 2.46\ \text{J/g}^\circ\text{C} \times 91.5^\circ\text{C}\)

Calculate \(q_1\):
\(75.8 \times 2.46 = 186.468\)
\(186.468 \times 91.5 = 17061.822\ \text{J} \approx 17.06\ \text{kJ}\)

Step 3: Calculate moles of ethanol:

\(n = \frac{m}{M} = \frac{75.8\ \text{g}}{46.07\ \text{g/mol}} \approx 1.645\ \text{mol}\)

Step 4: Calculate \(q_2\) (vaporization at boiling point):

\(q_2 = n\Delta H_{\text{vap}} = 1.645\ \text{mol} \times 38.56\ \text{kJ/mol}\)

Calculate \(q_2\):
\(1.645 \times 38.56 \approx 63.43\ \text{kJ}\)

Step 5: Total heat (\(q_{\text{total}}\)):

\(q_{\text{total}} = q_1 + q_2 = 17.06\ \text{kJ} + 63.43\ \text{kJ} \approx 80.49\ \text{kJ}\)

Wait, but maybe the problem assumes we only need to vaporize (if starting at boiling point), but the initial temp is \(-13.1^\circ\text{C}\), which is above the melting point (\(-114^\circ\text{C}\)), so ethanol is liquid. So we must heat liquid to boiling, then vaporize.

Wait, maybe I made a mistake in \(\Delta H_{\text{vap}}\) units. Let’s recheck:

If \(\Delta H_{\text{vap}} = 837\ \text{J/g}\) (alternative value, since \(38.56\ \text{kJ/mol} \div 46.07\ \text{g/mol} \approx 837\ \text{J/g}\)):

Then \(q_2 = 75.8\ \text{g} \times 837\ \text{J/g} = 63444.6\ \text{J} \approx 63.4\ \text{kJ}\)

Then \(q_1 = 75.8 \times 2.46 \times 91.5 = 17061\ \text{J} \approx 17.1\ \text{kJ}\)

Total \(q = 17.1 + 63.4 = 80.5\ \text{kJ}\) (rounded to 3 sig figs).

Final Answer:

\(\boldsymbol{80.5\ \text{kJ}}\) (or check with exact constants, but this is the process).

Answer:

To solve this, we need the following data (assumed or standard values for ethanol):

• Melting point of ethanol: \(-114^\circ\text{C}\) (but we start at \(-13.1^\circ\text{C}\), so we only heat liquid ethanol to its boiling point, then vaporize it? Wait, no—wait, the problem says "boil" ethanol, so we need to:

1. Heat liquid ethanol from \(-13.1^\circ\text{C}\) to its boiling point (\(T_b = 78.4^\circ\text{C}\)) using \(q_1 = mc\Delta T\).
2. Vaporize the liquid ethanol at its boiling point using \(q_2 = m\Delta H_{\text{vap}}\).

Step 1: Identify constants (standard values for ethanol):

• Specific heat of liquid ethanol, \(c = 2.46\ \text{J/g}^\circ\text{C}\)
• Molar mass of ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)): \(M = 46.07\ \text{g/mol}\)
• Enthalpy of vaporization, \(\Delta H_{\text{vap}} = 38.56\ \text{kJ/mol}\) (or \(38560\ \text{J/mol}\))

Step 2: Calculate \(q_1\) (heating liquid ethanol to boiling point):

\(\Delta T = T_b - T_{\text{initial}} = 78.4^\circ\text{C} - (-13.1^\circ\text{C}) = 91.5^\circ\text{C}\)
\(q_1 = mc\Delta T = 75.8\ \text{g} \times 2.46\ \text{J/g}^\circ\text{C} \times 91.5^\circ\text{C}\)

Calculate \(q_1\):
\(75.8 \times 2.46 = 186.468\)
\(186.468 \times 91.5 = 17061.822\ \text{J} \approx 17.06\ \text{kJ}\)

Step 3: Calculate moles of ethanol:

\(n = \frac{m}{M} = \frac{75.8\ \text{g}}{46.07\ \text{g/mol}} \approx 1.645\ \text{mol}\)

Step 4: Calculate \(q_2\) (vaporization at boiling point):

\(q_2 = n\Delta H_{\text{vap}} = 1.645\ \text{mol} \times 38.56\ \text{kJ/mol}\)

Calculate \(q_2\):
\(1.645 \times 38.56 \approx 63.43\ \text{kJ}\)

Step 5: Total heat (\(q_{\text{total}}\)):

\(q_{\text{total}} = q_1 + q_2 = 17.06\ \text{kJ} + 63.43\ \text{kJ} \approx 80.49\ \text{kJ}\)

Wait, but maybe the problem assumes we only need to vaporize (if starting at boiling point), but the initial temp is \(-13.1^\circ\text{C}\), which is above the melting point (\(-114^\circ\text{C}\)), so ethanol is liquid. So we must heat liquid to boiling, then vaporize.

Wait, maybe I made a mistake in \(\Delta H_{\text{vap}}\) units. Let’s recheck:

If \(\Delta H_{\text{vap}} = 837\ \text{J/g}\) (alternative value, since \(38.56\ \text{kJ/mol} \div 46.07\ \text{g/mol} \approx 837\ \text{J/g}\)):

Then \(q_2 = 75.8\ \text{g} \times 837\ \text{J/g} = 63444.6\ \text{J} \approx 63.4\ \text{kJ}\)

Then \(q_1 = 75.8 \times 2.46 \times 91.5 = 17061\ \text{J} \approx 17.1\ \text{kJ}\)

Total \(q = 17.1 + 63.4 = 80.5\ \text{kJ}\) (rounded to 3 sig figs).

Final Answer:

\(\boldsymbol{80.5\ \text{kJ}}\) (or check with exact constants, but this is the process).