Question

calculate the amount of heat needed to boil 116. g of ethanol (ch₃ch₂oh), beginning from a temperature of −75.3 °c. round your answer to 4 significant digits. also, be sure your answer contains a unit symbol.

Explanation:

To solve the problem of calculating the heat needed to boil ethanol, we need to consider three steps: heating the ethanol from \(-75.3^\circ \text{C}\) to its boiling point, vaporizing the ethanol at its boiling point. First, we need the specific heat capacity of liquid ethanol (\(c_{\text{liquid}}\)), the enthalpy of vaporization (\(\Delta H_{\text{vap}}\)), and the boiling point of ethanol (\(T_b = 78.4^\circ \text{C}\)).

Step 1: Calculate the moles of ethanol

The molar mass of ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)) is:

$$
LATEXBLOCK0
$$

Moles of ethanol (\(n\)):

$$ n = \frac{m}{M} = \frac{116 \, \text{g}}{46.068 \, \text{g/mol}} \approx 2.518 \, \text{mol} $$

Step 2: Heat the liquid ethanol from \(-75.3^\circ \text{C}\) to \(78.4^\circ \text{C}\)

The specific heat capacity of liquid ethanol (\(c_{\text{liquid}}\)) is \(2.46 \, \text{J/g}^\circ\text{C}\).
The temperature change (\(\Delta T\)):

$$ \Delta T = 78.4^\circ \text{C} - (-75.3^\circ \text{C}) = 153.7^\circ \text{C} $$

Heat required for heating (\(q_1\)):

$$ q_1 = mc\Delta T = 116 \, \text{g} \times 2.46 \, \text{J/g}^\circ\text{C} \times 153.7^\circ \text{C} $$

$$ q_1 \approx 116 \times 2.46 \times 153.7 \approx 44700 \, \text{J} = 44.7 \, \text{kJ} $$

Step 3: Vaporize the ethanol at its boiling point

The enthalpy of vaporization of ethanol (\(\Delta H_{\text{vap}}\)) is \(38.56 \, \text{kJ/mol}\).
Heat required for vaporization (\(q_2\)):

$$ q_2 = n\Delta H_{\text{vap}} = 2.518 \, \text{mol} \times 38.56 \, \text{kJ/mol} \approx 97.1 \, \text{kJ} $$

Step 4: Total heat required

$$ q_{\text{total}} = q_1 + q_2 = 44.7 \, \text{kJ} + 97.1 \, \text{kJ} = 141.8 \, \text{kJ} $$

Wait, let's check the calculations again for accuracy.

Recalculating Step 2:

$$ q_1 = 116 \, \text{g} \times 2.46 \, \text{J/g}^\circ\text{C} \times 153.7^\circ \text{C} $$

$$ 116 \times 2.46 = 285.36 $$

$$ 285.36 \times 153.7 \approx 285.36 \times 150 + 285.36 \times 3.7 = 42804 + 1055.832 = 43859.832 \, \text{J} \approx 43.86 \, \text{kJ} $$

Recalculating Step 3:

$$ q_2 = 2.518 \, \text{mol} \times 38.56 \, \text{kJ/mol} $$

$$ 2.518 \times 38.56 \approx 2.5 \times 38.56 + 0.018 \times 38.56 = 96.4 + 0.694 = 97.094 \, \text{kJ} $$

Total heat:

$$ q_{\text{total}} = 43.86 \, \text{kJ} + 97.094 \, \text{kJ} \approx 140.95 \, \text{kJ} \approx 141.0 \, \text{kJ} \, (\text{to 4 significant figures}) $$

Wait, maybe I made a mistake in the specific heat capacity or enthalpy of vaporization. Let's confirm the values:

• Specific heat capacity of liquid ethanol: \(2.46 \, \text{J/g}^\circ\text{C}\) (correct)
• Boiling point of ethanol: \(78.4^\circ \text{C}\) (correct)
• Enthalpy of vaporization of ethanol: \(38.56 \, \text{kJ/mol}\) (correct)

Moles of ethanol:

$$ n = \frac{116 \, \text{g}}{46.07 \, \text{g/mol}} \approx 2.518 \, \text{mol} \, (\text{correct}) $$

Temperature change:

$$ \Delta T = 78.4 - (-75.3) = 153.7^\circ \text{C} \, (\text{correct}) $$

Heat for heating:

$$ q_1 = 116 \times 2.46 \times 153.7 = 116 \times 378.102 = 43859.832 \, \text{J} = 43.86 \, \text{kJ} \, (\text{correct}) $$

Heat for vaporization:

$$ q_2 = 2.518 \times 38.56 = 97.09 \, \text{kJ} \, (\text{correct}) $$

Total heat:

$$ 43.86 + 97.09 = 140.95 \, \text{kJ} \approx 141.0 \, \text{kJ} $$

So the total heat required is approximately \(\boldsymbol{141.0 \, \text{kJ}}\) (to 4 significant figures).

Answer:

To solve the problem of calculating the heat needed to boil ethanol, we need to consider three steps: heating the ethanol from \(-75.3^\circ \text{C}\) to its boiling point, vaporizing the ethanol at its boiling point. First, we need the specific heat capacity of liquid ethanol (\(c_{\text{liquid}}\)), the enthalpy of vaporization (\(\Delta H_{\text{vap}}\)), and the boiling point of ethanol (\(T_b = 78.4^\circ \text{C}\)).

Step 1: Calculate the moles of ethanol

The molar mass of ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)) is:

$$
LATEXBLOCK0
$$

Moles of ethanol (\(n\)):

$$ n = \frac{m}{M} = \frac{116 \, \text{g}}{46.068 \, \text{g/mol}} \approx 2.518 \, \text{mol} $$

Step 2: Heat the liquid ethanol from \(-75.3^\circ \text{C}\) to \(78.4^\circ \text{C}\)

The specific heat capacity of liquid ethanol (\(c_{\text{liquid}}\)) is \(2.46 \, \text{J/g}^\circ\text{C}\).
The temperature change (\(\Delta T\)):

$$ \Delta T = 78.4^\circ \text{C} - (-75.3^\circ \text{C}) = 153.7^\circ \text{C} $$

Heat required for heating (\(q_1\)):

$$ q_1 = mc\Delta T = 116 \, \text{g} \times 2.46 \, \text{J/g}^\circ\text{C} \times 153.7^\circ \text{C} $$

$$ q_1 \approx 116 \times 2.46 \times 153.7 \approx 44700 \, \text{J} = 44.7 \, \text{kJ} $$

Step 3: Vaporize the ethanol at its boiling point

The enthalpy of vaporization of ethanol (\(\Delta H_{\text{vap}}\)) is \(38.56 \, \text{kJ/mol}\).
Heat required for vaporization (\(q_2\)):

$$ q_2 = n\Delta H_{\text{vap}} = 2.518 \, \text{mol} \times 38.56 \, \text{kJ/mol} \approx 97.1 \, \text{kJ} $$

Step 4: Total heat required

$$ q_{\text{total}} = q_1 + q_2 = 44.7 \, \text{kJ} + 97.1 \, \text{kJ} = 141.8 \, \text{kJ} $$

Wait, let's check the calculations again for accuracy.

Recalculating Step 2:

$$ q_1 = 116 \, \text{g} \times 2.46 \, \text{J/g}^\circ\text{C} \times 153.7^\circ \text{C} $$

$$ 116 \times 2.46 = 285.36 $$

$$ 285.36 \times 153.7 \approx 285.36 \times 150 + 285.36 \times 3.7 = 42804 + 1055.832 = 43859.832 \, \text{J} \approx 43.86 \, \text{kJ} $$

Recalculating Step 3:

$$ q_2 = 2.518 \, \text{mol} \times 38.56 \, \text{kJ/mol} $$

$$ 2.518 \times 38.56 \approx 2.5 \times 38.56 + 0.018 \times 38.56 = 96.4 + 0.694 = 97.094 \, \text{kJ} $$

Total heat:

$$ q_{\text{total}} = 43.86 \, \text{kJ} + 97.094 \, \text{kJ} \approx 140.95 \, \text{kJ} \approx 141.0 \, \text{kJ} \, (\text{to 4 significant figures}) $$

Wait, maybe I made a mistake in the specific heat capacity or enthalpy of vaporization. Let's confirm the values:

• Specific heat capacity of liquid ethanol: \(2.46 \, \text{J/g}^\circ\text{C}\) (correct)
• Boiling point of ethanol: \(78.4^\circ \text{C}\) (correct)
• Enthalpy of vaporization of ethanol: \(38.56 \, \text{kJ/mol}\) (correct)

Moles of ethanol:

$$ n = \frac{116 \, \text{g}}{46.07 \, \text{g/mol}} \approx 2.518 \, \text{mol} \, (\text{correct}) $$

Temperature change:

$$ \Delta T = 78.4 - (-75.3) = 153.7^\circ \text{C} \, (\text{correct}) $$

Heat for heating:

$$ q_1 = 116 \times 2.46 \times 153.7 = 116 \times 378.102 = 43859.832 \, \text{J} = 43.86 \, \text{kJ} \, (\text{correct}) $$

Heat for vaporization:

$$ q_2 = 2.518 \times 38.56 = 97.09 \, \text{kJ} \, (\text{correct}) $$

Total heat:

$$ 43.86 + 97.09 = 140.95 \, \text{kJ} \approx 141.0 \, \text{kJ} $$

So the total heat required is approximately \(\boldsymbol{141.0 \, \text{kJ}}\) (to 4 significant figures).